題意：

給你一棵樹，其實就是按照線段樹的方法建的樹，有兩種操作，將u到v的路徑上的邊權全部加w，然後詢問從u到v的路徑上的邊權和。

解法：

因為節點的編號很大，所以不能直接建樹，但是可以知道，這棵樹的最大深度也就是62層，所以我們可以用map來儲存邊權，或者把邊權放在點權上，就像樹鏈剖分那樣，每次操作時就可以暴力的向上更新，因為最多1000次操作，哪怕每次都是最深的節點，也最多向上63次，總體的複雜度是可以接受的

//  Created by  CQU_CST_WuErli
//  Copyright (c) 2016 CQU_CST_WuErli. All rights reserved.
 

//
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <string>
#include <vector>
#include <map>
#include <queue>
#include <stack>
 

#include <set>
#include <algorithm>
#include <sstream>
#define CLR(x) memset(x,0,sizeof(x))
#define OFF(x) memset(x,-1,sizeof(x))
#define MEM(x,a) memset((x),(a),sizeof(x))
#define BUG cout << "I am here" << endl
#define lookln(x) cout << #x << "=" << x << endl
 

#define SI(a) scanf("%d", &a)
#define SII(a,b) scanf("%d%d", &a, &b)
#define SIII(a,b,c) scanf("%d%d%d", &a, &b, &c)
const int INF_INT=0x3f3f3f3f;
const long long INF_LL=0x7f7f7f7f;
const int MOD=1e9+7;
const double eps=1e-10;
const double pi=acos(-1);
typedef long long  ll;
using namespace std;

int q;
ll logs[100];
map<ll, ll> sum;

void init() {
    logs[0] = 1LL;
    for (int i = 1; i <= 62;i++) {
        logs[i] = logs[i - 1] * 2 + 1;
    }
}

int get(ll x) {
    if (x == 1) return 0;
    else {
        for (int i = 1; i <= 62; i++) {
            if (x >= logs[i - 1] + 1 && x <= logs[i]) return i;
        }
    }
}

void change(ll u, ll v, ll w) {
    int du = get(u), dv = get(v);
    if (du > dv) {
        swap(du, dv);
        swap(u, v);
    }
    // v big
    for (int i = dv; i > du; i--) {
        sum[v] += w;
        v /= 2;
    }
    if (u == v) return;
    while (u != v) {
        sum[u] += w;
        sum[v] += w;
        u /= 2; v /= 2;
    }
}

ll query(ll u, ll v) {
    ll ret = 0;
    int du = get(u), dv = get(v);
    if (du > dv) {
        swap(du, dv);
        swap(u, v);
    }
    for (int i = dv; i > du; i--) {
        ret += sum[v];
        v /= 2;
    }
    if (u == v) return ret;
    while (u != v) {
        ret += sum[u];
        ret += sum[v];
        u /= 2, v /= 2;
    }
    return ret;
}

int main(int argc, char const *argv[]) {
#ifdef LOCAL
    freopen("C:\\Users\\john\\Desktop\\in.txt","r",stdin);
    // freopen("C:\\Users\\john\\Desktop\\out.txt","w",stdout);
#endif
    init();
    SI(q);
    for (int i = 1; i <= q; i++) {
        int op; SI(op);
        if (op == 1) {
            ll u, v, w;
            scanf("%I64d%I64d%I64d", &u, &v, &w);
            change(u, v, w);
        }
        else if (op == 2) {
            ll u, v;
            scanf("%I64d%I64d", &u, &v);
            printf("%I64d\n", query(u, v));
        }
    }
    return 0;
}