可以很顯然的發現，答案等於\(\sum{dist_{i,root}}-all(lca(l,l+1,l+2,,,,r))\)其中
我們可以使用線段樹取\([l,r]lca\)。
每一個點到根的和，我們考慮先離線，再\(access\)過程。每一次打通一條鏈，就要把這條鏈上，已經遍歷過的點到該鏈的鏈頭的距離給剪掉。在隨便使用一個數據結構維護字首和單點修改即可。因為離線，儲存R,每一次access後，一定是向後滿足條件的，所以每一次相當於詢問字首是否可行。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#include <iomanip>
#include<unordered_map>
#pragma GCC diagnostic error "-std=c++11"
#pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3)
#pragma GCC target("avx","sse2")
#define up(i,a,b)  for(int i=a;i<b;i++)
#define dw(i,a,b)  for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
	char ch = getchar(); ll x = 0, f = 1;
	while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 2e5 + 10;
int n, Q;
vector<int>vec[N];
vector<pir>ques[N];
int dep[N * 2], dfn[N * 2], rk[N * 2], st, dp[N * 2][30];
int ans[N];
struct sgt {
	int sum[N << 2];
	void pushup(int root) {
		sum[root] = sum[lrt] + sum[rrt];
	}
	void update(int l, int r, int root, int pos, int val) {
		if (l == r) {
			sum[root] += val; return;
		}
		int mid = (l + r) >> 1;
		if (pos <= mid)update(lson, pos, val);
		else update(rson, pos, val);
		pushup(root);
	}
	int query(int l, int r, int root, int lf, int rt) {
		if (lf > rt)return 0;
		if (lf <= l && r <= rt) { return sum[root]; }
		int mid = (l + r) >> 1;
		int ans = 0;
		if (lf <= mid)ans += query(lson, lf, rt);
		if (rt > mid)ans += query(rson, lf, rt);
		return ans;
	}
}T;
struct LCT {
	int ch[N][2], par[N], lz[N], val[N], sz[N];
	int ident(int rt, int fa) {
		return ch[fa][1] == rt;
	}
	void connect(int rt, int fa, int son) {
		par[rt] = fa;
		ch[fa][son] = rt;
	}
	bool isroot(int rt) {
		int f = par[rt];
		return ch[f][1] != rt && ch[f][0] != rt;
	}
	void rotate(int rt) {
		int f = par[rt]; int ff = par[f]; int k = ident(rt, f);
		connect(ch[rt][k ^ 1], f, k);
		par[rt] = ff;
		if (!isroot(f))ch[ff][ident(f, ff)] = rt;
		connect(f, rt, k ^ 1);
		pushup(f); pushup(rt);
	}
	void pushup(int rt) {
		sz[rt] = sz[ch[rt][0]] + sz[ch[rt][1]] + 1;
	}
	void pushdown(int rt) {
		if (lz[rt]) {
			int ls = ch[rt][0];
			int rs = ch[rt][1];
			if (ls)val[ls] = lz[ls] = lz[rt];
			if (rs)val[rs] = lz[rs] = lz[rt];
			lz[rt] = 0;
		}
	}
	void pushall(int rt) {
		if (!isroot(rt))pushall(par[rt]);
		pushdown(rt);
	}
	void splay(int x) {
		pushall(x);
		while (!isroot(x)) {
			int f = par[x]; int ff = par[f];
			if (!isroot(f))ident(f, ff) ^ ident(x, f) ? rotate(x) : rotate(f);
			rotate(x);
		}
	}
	void access(int x, int col) {
		int y;
		for (y = 0; x; y = x, x = par[x]) {
			splay(x);
			if (val[x])
				T.update(1, n, 1, val[x], -sz[ch[x][0]] - 1);
			T.update(1, n, 1, col, sz[ch[x][0]] + 1);
			//cout <<val[x]<<" "<< T.query(1, n, 1, 1, val[x]) << " ";
			ch[x][1] = y;
			pushup(x);
		}
		val[y] = lz[y] = col;
	}
}lct;
void dfs1(int u, int fa, int d) {
	lct.par[u] = fa;
	dfn[++st] = u; dep[st] = d;
	rk[u] = st;
	for (auto k : vec[u]) {
		if (k == fa)continue;
		dfs1(k, u, d + 1);
		dfn[++st] = u;
		dep[st] = d;
		rk[u] = st;
	}
}
void initst() {
	upd(i, 0, 2 * n)dp[i][0] = i;
	upd(j, 1, 30) {
		for (int i = 1; i + (1 << j) - 1 <= 2 * n; i++) {
			int t1 = dp[i][j - 1]; int t2 = dp[i + (1 << (j - 1))][j - 1];
			dp[i][j] = dep[t1] < dep[t2] ? t1 : t2;
		}
	}
}
int lca(int x, int y) {
	x = rk[x], y = rk[y];
	if (x > y)swap(x, y);
	int k = int(log2(y - x + 1));
	int t1 = dp[x][k]; int t2 = dp[y - (1 << k) + 1][k];
	return dep[t1] < dep[t2] ? dfn[t1] : dfn[t2];
}
struct LCAT {
	int LCA[N << 2];
	void pushup(int root) {
		LCA[root] = lca(LCA[lrt], LCA[rrt]);
	}
	void build(int l, int r, int root) {
		if (l == r) {
			LCA[root] = l; return;
		}
		int mid = (l + r) >> 1;
		build(lson); build(rson); pushup(root);
	}
	int query(int l, int r, int root, int lf, int rt) {
		if (lf <= l && r <= rt) {
			return LCA[root];
		}
		int mid = (l + r) >> 1;
		int temp = -1;
		if (lf <= mid)temp = query(lson, lf, rt);
		if (rt > mid) {
			if (temp == -1)temp = query(rson, lf, rt);
			else temp = lca(temp, query(rson, lf, rt));
		}
		return temp;
	}
}LCAT;
int main() {
	n = read(); Q = read();
	int u, v;
	upd(i, 1, n - 1) {
		u = read(), v = read();
		vec[u].push_back(v);
		vec[v].push_back(u);
	}
	upd(i, 1, Q) {
		u = read(), v = read();
		ques[v].push_back({ u,i });
	}
	dfs1(1, 0, 0);
	initst();
	upd(i, 1, n) {
		 lct.sz[i] = 1;
	}
	LCAT.build(1, n, 1);
	//cout << LCAT.query(1, n, 1, 1, 2);
	upd(i, 1, n) {
		lct.access(i, i);
		////cout << T.query(1, n, 1, 1, i);
		//upd(j, 1, i) {
		//	cout << T.query(1, n, 1, 1, j) << "  ";
		//}cout << endl;
		for (auto k : ques[i]) {
			ans[k.second] = T.query(1, n, 1, k.first, i) - dep[rk[LCAT.query(1, n, 1, k.first, i)]] - 1;
		}
	}
	upd(i, 1, Q)printf("%d\n", ans[i]);
	return 0;
}