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;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
const long long N = 2e5 + 5, inf = 1e18;
vector<int> g[N];
long long a[N], p[N], b[N], tin[N], tout[N], tmr;
void dfs(int u, int par) {
  tin[u] = ++tmr;
  for (int v : g[u])
    if (v != par) {
      p[v] += p[u];
      dfs(v, u);
    }
  tout[u] = tmr;
}
bool is_anc(int u, int v) { return tin[u] <= tin[v] and tout[u] >= tout[v]; }
struct T {
  long long t[4 * N], lazy[4 * N];
  void prop(int i, int st, int en) {
    if (lazy[i]) {
      t[i] += lazy[i];
      if (st != en) {
        lazy[2 * i] += lazy[i];
        lazy[2 * i + 1] += lazy[i];
      }
      lazy[i] = 0;
    }
  }
  void update(int i, int st, int en, int l, int r, long long v) {
    prop(i, st, en);
    if (st > r or en < l) {
      return;
    }
    if (st >= l and en <= r) {
      lazy[i] = v;
      prop(i, st, en);
      return;
    }
    int mid = (st + en) / 2;
    update(2 * i, st, mid, l, r, v);
    update(2 * i + 1, mid + 1, en, l, r, v);
    t[i] = min(t[2 * i], t[2 * i + 1]);
  }
  long long query(int i, int st, int en, int l, int r) {
    prop(i, st, en);
    if (st > r or en < l) {
      return inf;
    }
    if (st >= l and en <= r) {
      return t[i];
    }
    int mid = (st + en) / 2;
    long long x = query(2 * i, st, mid, l, r);
    long long y = query(2 * i + 1, mid + 1, en, l, r);
    return min(x, y);
  }
} t[2];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n >> q;
  int e[2 * n];
  for (int i = 1; i <= 2 * n - 2; i++) {
    int x, y, w;
    cin >> x >> y >> w;
    if (i <= n - 1) {
      g[x].push_back(y);
      a[y] = p[y] = w;
      e[i] = y;
    } else {
      b[x] = w;
      e[i] = x;
    }
  }
  dfs(1, 0);
  for (int i = 1; i <= n; i++) {
    t[0].update(1, 1, n, tin[i], tin[i], p[i]);
    t[1].update(1, 1, n, tin[i], tin[i], p[i] + b[i]);
  }
  while (q--) {
    int ty, x, y;
    cin >> ty >> x >> y;
    if (ty == 1) {
      int u = e[x];
      if (x <= n - 1) {
        t[0].update(1, 1, n, tin[u], tout[u], y - a[u]);
        t[1].update(1, 1, n, tin[u], tout[u], y - a[u]);
        a[u] = y;
      } else {
        t[1].update(1, 1, n, tin[u], tin[u], y - b[u]);
        b[u] = y;
      }
    }
    if (ty == 2) {
      long long ans, tmp1, tmp2;
      tmp1 = t[0].query(1, 1, n, tin[x], tin[x]);
      tmp2 = t[0].query(1, 1, n, tin[y], tin[y]);
      ans = t[1].query(1, 1, n, tin[x], tout[x]) - tmp1 + tmp2;
      if (is_anc(x, y)) {
        ans = min(ans, tmp2 - tmp1);
      }
      cout << ans << "\n";
    }
  }
}