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;
struct edge {
  int y, z, nxt;
} e[400010];
long long tree[800010];
int laz[800010];
int head[200010];
long long dis[200010];
int eg[400010][3];
int num[200010];
int siz[200010];
int dep[200010];
int to[200010];
int ed[200010];
int n, m, cnt, idx;
void add(int x, int y, int z) {
  cnt++;
  e[cnt].y = y;
  e[cnt].z = z;
  e[cnt].nxt = head[x];
  head[x] = cnt;
}
void dfs(int x, int f) {
  siz[x] = 1;
  num[x] = ++idx;
  to[idx] = x;
  for (int i = head[x]; i; i = e[i].nxt) {
    int y = e[i].y, z = e[i].z;
    if (y == f) continue;
    dis[y] = dis[x] + z;
    dep[y] = dep[x] + 1;
    dfs(y, x);
    siz[x] += siz[y];
  }
}
void build(int id, int l, int r) {
  if (l == r) {
    int x = to[l];
    tree[id] = dis[x] + eg[ed[x]][2];
    return;
  }
  int mid = (l + r) >> 1;
  build(id << 1, l, mid);
  build(id << 1 | 1, mid + 1, r);
  tree[id] = min(tree[id << 1], tree[id << 1 | 1]);
}
void pushdown(int id) {
  if (laz[id]) {
    tree[id << 1] += laz[id];
    tree[id << 1 | 1] += laz[id];
    laz[id << 1] += laz[id];
    laz[id << 1 | 1] += laz[id];
    laz[id] = 0;
  }
}
void update(int id, int l, int r, int ul, int ur, int x) {
  if (ul <= l && r <= ur) {
    tree[id] += x;
    laz[id] += x;
    return;
  }
  pushdown(id);
  int mid = (l + r) >> 1;
  if (ul <= mid) update(id << 1, l, mid, ul, ur, x);
  if (ur > mid) update(id << 1 | 1, mid + 1, r, ul, ur, x);
  tree[id] = min(tree[id << 1], tree[id << 1 | 1]);
}
long long query(int id, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr) return tree[id];
  pushdown(id);
  int mid = (l + r) >> 1;
  long long ans = 2e17;
  if (ql <= mid) ans = min(ans, query(id << 1, l, mid, ql, qr));
  if (qr > mid) ans = min(ans, query(id << 1 | 1, mid + 1, r, ql, qr));
  return ans;
}
int main() {
  int op, x, y;
  scanf("%d%d", &n, &m);
  for (int i = 1; i < 2 * n - 1; i++) {
    scanf("%d%d%d", &eg[i][0], &eg[i][1], &eg[i][2]);
    if (i < n)
      add(eg[i][0], eg[i][1], eg[i][2]), add(eg[i][1], eg[i][0], eg[i][2]);
    else
      ed[eg[i][0]] = i;
  }
  dfs(1, -1);
  build(1, 1, n);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1) {
      int u = eg[x][0];
      int v = eg[x][1];
      int w = eg[x][2];
      if (x < n)
        update(1, 1, n, num[v], num[v] + siz[v] - 1, y - w);
      else
        update(1, 1, n, num[u], num[u], y - w);
      eg[x][2] = y;
    } else {
      if (num[x] <= num[y] && num[y] < num[x] + siz[x])
        printf("%lld\n", query(1, 1, n, num[y], num[y]) - eg[ed[y]][2] -
                             query(1, 1, n, num[x], num[x]) + eg[ed[x]][2]);
      else
        printf("%lld\n", query(1, 1, n, num[x], num[x] + siz[x] - 1) -
                             query(1, 1, n, num[x], num[x]) + eg[ed[x]][2] +
                             query(1, 1, n, num[y], num[y]) - eg[ed[y]][2]);
    }
  }
}