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 maxn = 4e5 + 10;
int a[maxn], b[maxn], c[maxn], d[maxn], f[maxn][22], dep[maxn], lg[maxn];
struct node {
  int to, cost;
};
vector<node> G[maxn];
int n, Q, in[maxn], out[maxn], re[maxn], tol;
long long s[maxn], tree[maxn << 2], lazy[maxn << 2];
void build(int l, int r, int rt) {
  if (l == r) {
    tree[rt] = s[re[l]] + d[re[l]];
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, rt << 1);
  build(mid + 1, r, rt << 1 | 1);
  tree[rt] = min(tree[rt << 1], tree[rt << 1 | 1]);
}
void dfs(int v, int dist) {
  in[v] = ++tol;
  re[tol] = v;
  dep[v] = dist;
  for (int i = 1; (1 << i) <= dep[v]; i++) f[v][i] = f[f[v][i - 1]][i - 1];
  for (int i = 0; i < G[v].size(); i++) {
    int to = G[v][i].to;
    s[to] = s[v] + G[v][i].cost;
    f[to][0] = v;
    dfs(to, dist + 1);
  }
  out[v] = tol;
}
void push_down(int rt, int l, int r) {
  if (lazy[rt]) {
    int mid = (l + r) / 2;
    tree[rt << 1] += lazy[rt];
    tree[rt << 1 | 1] += lazy[rt];
    lazy[rt << 1] += lazy[rt];
    lazy[rt << 1 | 1] += lazy[rt];
    lazy[rt] = 0;
  }
}
void update(int L, int R, int C, int l, int r, int rt) {
  if (L <= l && r <= R) {
    tree[rt] += C;
    lazy[rt] += C;
    return;
  }
  push_down(rt, l, r);
  int mid = (l + r) / 2;
  if (L <= mid) update(L, R, C, l, mid, rt << 1);
  if (R > mid) update(L, R, C, mid + 1, r, rt << 1 | 1);
  tree[rt] = min(tree[rt << 1], tree[rt << 1 | 1]);
}
long long query(int L, int R, int l, int r, int rt) {
  if (l == r) return tree[rt];
  push_down(rt, l, r);
  int mid = (l + r) / 2;
  if (L <= mid && R > mid)
    return min(query(L, R, l, mid, rt << 1),
               query(L, R, mid + 1, r, rt << 1 | 1));
  else if (L <= mid)
    return query(L, R, l, mid, rt << 1);
  else
    return query(L, R, mid + 1, r, rt << 1 | 1);
}
bool check(int x, int y) {
  if (dep[x] > dep[y]) return 0;
  while (dep[x] < dep[y]) {
    int h = lg[dep[y] - dep[x]];
    y = f[y][h];
  }
  return x == y;
}
int main() {
  for (int i = 2; i < maxn; i++) lg[i] = lg[i / 2] + 1;
  scanf("%d%d", &n, &Q);
  for (int i = 1; i <= 2 * n - 2; i++) {
    scanf("%d%d%d", &a[i], &b[i], &c[i]);
    if (i <= n - 1)
      G[a[i]].push_back((node){b[i], c[i]});
    else
      d[a[i]] = c[i];
  }
  dfs(1, 1);
  build(1, n, 1);
  while (Q--) {
    int op, x, y;
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1) {
      if (x <= n - 1)
        update(in[b[x]], out[b[x]], y - c[x], 1, n, 1), c[x] = y;
      else
        update(in[a[x]], in[a[x]], y - d[a[x]], 1, n, 1), d[a[x]] = y;
    } else if (x == y)
      printf("0\n");
    else if (check(x, y))
      printf("%lld\n", query(in[y], in[y], 1, n, 1) -
                           query(in[x], in[x], 1, n, 1) - d[y] + d[x]);
    else {
      long long ans =
          query(in[x], out[x], 1, n, 1) - query(in[x], in[x], 1, n, 1) + d[x];
      ans += query(in[y], in[y], 1, n, 1) - d[y];
      printf("%lld\n", ans);
    }
  }
  return 0;
}