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;
inline int read() {
  int x = 0, t = 1, c;
  while (!isdigit(c = getchar()))
    if (c == '-') t = -1;
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  return x * t;
}
int n, q;
int first[200020], targ[200020], nxt[200020], dist[200020], dfv[200020],
    dfe[200020], tp[200020], port[200020], cntd = 1, cnte = 1;
void AddEdge(int u, int v, int c) {
  targ[cnte] = v;
  dist[cnte] = c;
  nxt[cnte] = first[u];
  first[u] = cnte++;
}
long long addv[200020 << 2], minv[200020 << 2];
void maintain(int o) {
  int lc = o << 1, rc = lc | 1;
  minv[o] = min(minv[lc] + addv[lc], minv[rc] + addv[rc]);
}
void Add(int o, int L, int R, const int &ql, const int &qr, const int &v) {
  if (ql <= L && R <= qr) {
    addv[o] += v;
  } else {
    int m = L + R >> 1, lc = o << 1, rc = lc | 1;
    if (ql <= m) Add(lc, L, m, ql, qr, v);
    if (m < qr) Add(rc, m + 1, R, ql, qr, v);
    maintain(o);
  }
}
long long Query(int o, int L, int R, const int &ql, const int &qr,
                long long added = 0) {
  added += addv[o];
  if (ql <= L && R <= qr) {
    return minv[o] + added;
  } else {
    int m = L + R >> 1, lc = o << 1, rc = lc | 1;
    long long ans = (1ll << 63ll) - 1ll;
    if (ql <= m) ans = Query(lc, L, m, ql, qr, added);
    if (m < qr) ans = min(ans, Query(rc, m + 1, R, ql, qr, added));
    return ans;
  }
}
void DFS(int x) {
  dfv[x] = cntd++;
  for (int i = first[x]; i; i = nxt[i]) DFS(targ[i]);
  dfe[x] = cntd - 1;
}
int main() {
  n = read(), q = read();
  for (int i = 1; i < n; i++) {
    int a = read(), b = read(), c = read();
    AddEdge(a, b, c);
  }
  for (int i = 1; i < n; i++) {
    int a = read(), b = read(), c = read();
    tp[a] = c;
    port[i] = a;
  }
  DFS(1);
  for (int i = 1; i < cnte; i++)
    Add(1, 1, n, dfv[targ[i]], dfe[targ[i]], dist[i]);
  for (int i = 1; i <= n; i++) Add(1, 1, n, dfv[i], dfv[i], tp[i]);
  while (q--) {
    int o = read(), a = read(), b = read();
    if (o == 1) {
      if (a >= n) {
        a -= n - 1;
        a = port[a];
        Add(1, 1, n, dfv[a], dfv[a], b - tp[a]);
        tp[a] = b;
      } else {
        Add(1, 1, n, dfv[targ[a]], dfe[targ[a]], b - dist[a]);
        dist[a] = b;
      }
    } else {
      long long deepa = Query(1, 1, n, dfv[a], dfv[a]) - tp[a],
                Tp = Query(1, 1, n, dfv[a], dfe[a]) - deepa,
                deepb = Query(1, 1, n, dfv[b], dfv[b]) - tp[b];
      if (dfv[b] >= dfv[a] && dfe[b] <= dfe[a]) {
        printf("%I64d\n", deepb - deepa);
      } else {
        printf("%I64d\n", deepb + Tp);
      }
    }
  }
}