Problem: You are given a directed weighted graph with n nodes and 2n - 2 edges. The nodes are labeled from 1 to n, while the edges are labeled from 1 to 2n - 2. The graph's edges can be split into two parts.

  * The first n - 1 edges will form a rooted spanning tree, with node 1 as the root. All these edges will point away from the root. 
  * The last n - 1 edges will be from node i to node 1, for all 2 ≤ i ≤ n. 



You are given q queries. There are two types of queries

  * 1 i w: Change the weight of the i-th edge to w
  * 2 u v: Print the length of the shortest path between nodes u to v



Given these queries, print the shortest path lengths.

Input

The first line of input will contain two integers n, q (2 ≤ n, q ≤ 200 000), the number of nodes, and the number of queries, respectively.

The next 2n - 2 integers will contain 3 integers ai, bi, ci, denoting a directed edge from node ai to node bi with weight ci.

The first n - 1 of these lines will describe a rooted spanning tree pointing away from node 1, while the last n - 1 of these lines will have bi = 1.

More specifically, 

  * The edges (a1, b1), (a2, b2), ... (an - 1, bn - 1) will describe a rooted spanning tree pointing away from node 1. 
  * bj = 1 for n ≤ j ≤ 2n - 2. 
  * an, an + 1, ..., a2n - 2 will be distinct and between 2 and n. 



The next q lines will contain 3 integers, describing a query in the format described in the statement.

All edge weights will be between 1 and 106.

Output

For each type 2 query, print the length of the shortest path in its own line.

Example

Input

5 9
1 3 1
3 2 2
1 4 3
3 5 4
5 1 5
3 1 6
2 1 7
4 1 8
2 1 1
2 1 3
2 3 5
2 5 2
1 1 100
2 1 3
1 8 30
2 4 2
2 2 4


Output

0
1
4
8
100
132
10

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAX = 5e5 + 5;
const long long INF = 1LL << 60;
int a[MAX], b[MAX], c[MAX];
long long segtree[4 * MAX], lazy[4 * MAX];
int n, q;
void push(int p, int l, int r) {
  if (lazy[p] != 0) {
    segtree[p] += lazy[p];
    if (l != r) {
      lazy[p * 2] += lazy[p];
      lazy[p * 2 + 1] += lazy[p];
    }
    lazy[p] = 0;
  }
}
void update(int p, int l, int r, int i, int j, int val) {
  push(p, l, r);
  if (i > r || l > j) {
    return;
  } else if (l >= i && j >= r) {
    segtree[p] += val;
    if (l != r) {
      lazy[p * 2] += val;
      lazy[p * 2 + 1] += val;
    }
    return;
  }
  int mid = (l + r) / 2;
  update(p * 2, l, mid, i, j, val);
  update(p * 2 + 1, mid + 1, r, i, j, val);
  segtree[p] = min(segtree[p * 2], segtree[p * 2 + 1]);
}
long long query(int p, int l, int r, int i, int j) {
  push(p, l, r);
  if (i > r || l > j) {
    return INF;
  } else if (l >= i && j >= r) {
    return segtree[p];
  }
  int mid = (l + r) / 2;
  return min(query(p * 2, l, mid, i, j), query(p * 2 + 1, mid + 1, r, i, j));
}
vector<int> adj[MAX];
int st[MAX], en[MAX], weight[MAX];
int dfs_c = 1;
void dfs(int source) {
  st[source] = dfs_c++;
  for (auto &each : adj[source]) {
    dfs(each);
  }
  en[source] = dfs_c - 1;
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = int(1); i < int(n - 1 + 1); i++) {
    scanf("%d %d %d", a + i, b + i, c + i);
    adj[a[i]].emplace_back(b[i]);
  }
  dfs(1);
  for (int i = int(n); i < int(2 * n - 2 + 1); i++) {
    scanf("%d %d %d", a + i, b + i, c + i);
    weight[a[i]] = c[i];
    adj[a[i]].emplace_back(b[i]);
  }
  for (int i = int(1); i < int(n - 1 + 1); i++) {
    update(1, 1, n, st[b[i]], en[b[i]], c[i]);
  }
  for (int i = int(n); i < int(2 * n - 2 + 1); i++) {
    update(1, 1, n, st[a[i]], st[a[i]], c[i]);
  }
  while (q--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int idx, w;
      scanf("%d %d", &idx, &w);
      if (idx <= n - 1) {
        update(1, 1, n, st[b[idx]], en[b[idx]], -c[idx]);
        c[idx] = w;
        update(1, 1, n, st[b[idx]], en[b[idx]], c[idx]);
      } else {
        update(1, 1, n, st[a[idx]], st[a[idx]], -c[idx]);
        c[idx] = w;
        weight[a[idx]] = c[idx];
        update(1, 1, n, st[a[idx]], st[a[idx]], c[idx]);
      }
    } else {
      int u, v;
      scanf("%d %d", &u, &v);
      if (st[u] <= st[v] && en[v] <= en[u]) {
        long long ru = query(1, 1, n, st[u], st[u]) - weight[u];
        long long rv = query(1, 1, n, st[v], st[v]) - weight[v];
        printf("%lld\n", rv - ru);
      } else {
        long long ur = query(1, 1, n, st[u], en[u]) -
                       (query(1, 1, n, st[u], st[u]) - weight[u]);
        long long rv = query(1, 1, n, st[v], st[v]) - weight[v];
        printf("%lld\n", ur + rv);
      }
    }
  }
  return 0;
}