E. New Year Tree time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output

The New Year holidays are over, but Resha doesn't want to throw away the New Year tree. He invited his best friends Kerim and Gural to help him to redecorate the New Year tree.

The New Year tree is an undirected tree withnvertices and root in the vertex1.

You should process the queries of the two types:

1. Change the colours of all vertices in the subtree of the vertexvto the colourc.
2. Find the number of different colours in the subtree of the vertexv.

Input

The first line contains two integersn, m(1 ≤ n, m ≤ 4·10⁵) — the number of vertices in the tree and the number of the queries.

The second line containsn

integersc_i(1 ≤ c_i ≤ 60) — the colour of thei-th vertex.

Each of the nextn - 1lines contains two integersx_j, y_j(1 ≤ x_j, y_j ≤ n) — the vertices of thej-th edge. It is guaranteed that you are given correct undirected tree.

The lastmlines contains the description of the queries. Each description starts with the integert

_k(1 ≤ t_k ≤ 2) — the type of thek-th query. For the queries of the first type then follows two integersv_k, c_k(1 ≤ v_k ≤ n, 1 ≤ c_k ≤ 60) — the number of the vertex whose subtree will be recoloured with the colourc_k. For the queries of the second type then follows integerv_k(1 ≤ v_k ≤ n) — the number of the vertex for which subtree you should find the number of different colours.

Output

For each query of the second type print the integera— the number of different colours in the subtree of the vertex given in the query.

Each of the numbers should be printed on a separate line in order of query appearing in the input.

Examples input

7 10
1 1 1 1 1 1 1
1 2
1 3
1 4
3 5
3 6
3 7
1 3 2
2 1
1 4 3
2 1
1 2 5
2 1
1 6 4
2 1
2 2
2 3

output

2
3
4
5
1
2

input

23 30
1 2 2 6 5 3 2 1 1 1 2 4 5 3 4 4 3 3 3 3 3 4 6
1 2
1 3
1 4
2 5
2 6
3 7
3 8
4 9
4 10
4 11
6 12
6 13
7 14
7 15
7 16
8 17
8 18
10 19
10 20
10 21
11 22
11 23
2 1
2 5
2 6
2 7
2 8
2 9
2 10
2 11
2 4
1 12 1
1 13 1
1 14 1
1 15 1
1 16 1
1 17 1
1 18 1
1 19 1
1 20 1
1 21 1
1 22 1
1 23 1
2 1
2 5
2 6
2 7
2 8
2 9
2 10
2 11
2 4

output

6
1
3
3
2
1
2
3
5
5
1
2
2
1
1
1
2
3

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1,class T2>inline void gmax(T1 &a,T2 b){if(b>a)a=b;}
template <class T1,class T2>inline void gmin(T1 &a,T2 b){if(b<a)a=b;}
const int N=4e5+10,M=0,Z=1e9+7,ms63=0x3f3f3f3f;
int n,m,x,y;
int col[N];
int first[N], id; int w[N << 1], nxt[N << 1]; bool vis[N];
int in[N], out[N]; int p[N]; int tim;
LL b[64];
LL C;
struct A
{
	int l,r;
	LL flag;
	LL c;
}a[1<<20];
void ins(int x,int y)
{
	++id;
	w[id]=y;
	nxt[id]=first[x];
	first[x]=id;
}
void dfs(int x)
{
	vis[x]=1;
	in[x]=++tim;
	p[tim]=col[x];
	for(int z=first[x];z;z=nxt[z])
	{
		int y=w[z];
		if(vis[y])continue;
		dfs(y);
	}
	out[x]=tim;
}
void pushdown(int o)
{
	if(a[o].flag)
	{
		a[ls].flag=a[rs].flag=a[o].flag;
		a[ls].c |= a[o].flag;
		a[rs].c |= a[o].flag;
		a[o].flag=0;
	}
}
void pushup(int o)
{
	a[o].c=a[ls].c|a[rs].c;
}
void build(int o,int l,int r)
{
	a[o].l = l;
	a[o].r = r;
	a[o].flag = 0;
	if(l==r)
	{
		a[o].c|=b[ p[l] ];
		return;
	}
	int mid=(l+r)>>1;
	build(ls,l,mid);
	build(rs,mid+1,r);
	pushup(o);
	return;
}
void change(int o,int l,int r,LL v)
{
	if(a[o].l==l&&a[o].r==r)
	{
		a[o].flag = a[o].c = v;
		return;
	}
	pushdown(o);
	int mid=(a[o].l+a[o].r)>>1;
	if(r<=mid)change(ls,l,r,v);
	else if(l>mid)change(rs,l,r,v);
	else
	{
		change(ls,l,mid,v);
		change(rs,mid+1,r,v);
	}
	pushup(o);
}
void update(int o,int l,int r)
{
	if(a[o].flag)
	{
		C |= a[o].flag;
		return;
	}
	if(a[o].l==l&&a[o].r==r)
	{
		C |= a[o].c;
		return;
	}
	int mid=(a[o].l+a[o].r)>>1;
	if(r<=mid)update(ls,l,r);
	else if(l>mid)update(rs,l,r);
	else
	{
		update(ls,l,mid);
		update(rs,mid+1,r);
	}
	pushup(o);
}
int main()
{
	for (int i = 0; i <= 60; ++i)b[i] = 1ll << i;
	while(~scanf("%d%d",&n,&m))
	{
		memset(first,0,n+2<<2);id=1;
		for(int i=1;i<=n;++i)scanf("%d",&col[i]);
		for(int i=1;i<n;++i)
		{
			scanf("%d%d",&x,&y);
			ins(x,y);
			ins(y,x);
		}
		MS(vis,0);
		tim=0;
		dfs(1);
		build(1,1,n);
		for(int i=1;i<=m;++i)
		{
			int o;
			int p,v;
			scanf("%d",&o);
			if(o==1)
			{
				scanf("%d%d",&p,&v);
				change(1,in[p],out[p],b[v]);
			}
			else
			{
				scanf("%d",&p);
				C = 0;
				update(1,in[p],out[p]);
				printf("%d\n", __builtin_popcountll(C));
			}
		}
	}
	return 0;
}
/*
【題意】
給你一棵樹，有n（4e5）個節點，每個節點有一個顏色，顏色範圍在[1,60]
我們有兩種操作。
操作1：把節點p的子樹都改為顏色v
操作2：查詢以p為根的子樹有多少種不同的顏色。

【型別】
線段樹
dfs序

【分析】
我們先得到in和out的dfs序，
然後建線段樹，每個節點開60個bool陣列，表示顏色數
（或者用LL來表示）

然後對於修改操作，使用flag打標記
對於查詢操作，做全域性或運算，然後用__builtin_popcountll()求1的個數即可

【時間複雜度&&優化】
O(mlogn)

*/