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;
using ll = long long;
const int N = 2e5L + 11, M = N * 2;
int a[M], b[M], c[M];
int back[M];
struct SegTree {
  ll flag[N * 8];
  ll val[N * 4];
  void push(int x) {
    val[x] += flag[x];
    flag[2 * x] += flag[x];
    flag[2 * x + 1] += flag[x];
    flag[x] = 0;
  }
  void change(int l, int r, int ql, int qr, int c, int x = 1) {
    push(x);
    if (ql == l && qr == r) {
      flag[x] = c;
      push(x);
    } else {
      if (qr <= (l + r) / 2) {
        change(l, (l + r) / 2, ql, qr, c, 2 * x);
        push(2 * x + 1);
      } else if (ql >= (l + r) / 2) {
        push(2 * x);
        change((l + r) / 2, r, ql, qr, c, 2 * x + 1);
      } else {
        change(l, (l + r) / 2, ql, (l + r) / 2, c, 2 * x);
        change((l + r) / 2, r, (l + r) / 2, qr, c, 2 * x + 1);
      }
      val[x] = min(val[2 * x], val[2 * x + 1]);
    }
  }
  ll query(int l, int r, int ql, int qr, int x = 1) {
    push(x);
    if (ql == l && qr == r) return val[x];
    if (qr <= (l + r) / 2)
      return query(l, (l + r) / 2, ql, qr, 2 * x);
    else if (ql >= (l + r) / 2)
      return query((l + r) / 2, r, ql, qr, 2 * x + 1);
    else
      return min(query(l, (l + r) / 2, ql, (l + r) / 2, 2 * x),
                 query((l + r) / 2, r, (l + r) / 2, qr, 2 * x + 1));
  }
} st;
vector<int> e[N];
int cnt = 0;
int in[N], out[N], id[N];
void dfs(int x) {
  id[x] = ++cnt;
  in[x] = cnt;
  for (int i : e[x]) dfs(i);
  out[x] = cnt;
}
int n;
ll query(int x) { return st.query(1, n + 1, id[x], id[x] + 1) - back[x]; }
int main() {
  int q;
  cin >> 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)
      e[a[i]].push_back(b[i]);
    else
      back[a[i]] = c[i];
  }
  dfs(1);
  for (int i = 1; i <= n - 1; i++)
    st.change(1, n + 1, in[b[i]], out[b[i]] + 1, c[i]);
  for (int i = 2; i <= n; i++) st.change(1, n + 1, id[i], id[i] + 1, back[i]);
  while (q--) {
    int t, u, v;
    scanf("%d%d%d", &t, &u, &v);
    if (t == 1) {
      int i = u, w = v;
      if (i <= n - 1) {
        st.change(1, n + 1, in[b[i]], out[b[i]] + 1, w - c[i]);
        c[i] = w;
      } else {
        st.change(1, n + 1, id[a[i]], id[a[i]] + 1, w - c[i]);
        back[a[i]] = w;
        c[i] = w;
      }
    } else {
      ll ans = st.query(1, n + 1, in[u], out[u] + 1) - query(u) + query(v);
      if (id[v] >= in[u] && id[v] <= out[u])
        ans = min(ans, query(v) - query(u));
      printf("%lld\n", ans);
    }
  }
}