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 N = 4 * 1e5 + 69;
long long it_from_root[4 * N], it_to_root[4 * N], lz_from_root[4 * N],
    lz_to_root[4 * N];
tuple<int, int, int> edge[N];
int edge_to_root[N], n, q, p[N], l[N], r[N], timer;
vector<pair<int, int> > a[N];
void lazy(int p, int l, int r) {
  it_from_root[p] += lz_from_root[p];
  it_to_root[p] += lz_to_root[p];
  if (l != r) {
    lz_from_root[p * 2] += lz_from_root[p];
    lz_from_root[p * 2 + 1] += lz_from_root[p];
    lz_to_root[p * 2] += lz_to_root[p];
    lz_to_root[p * 2 + 1] += lz_to_root[p];
  }
  lz_from_root[p] = lz_to_root[p] = 0;
}
void update(int p, int l, int r, int ql, int qr, long long val_from_root,
            long long val_to_root) {
  lazy(p, l, r);
  if (ql > r || l > qr) return;
  if (ql <= l && r <= qr) {
    it_from_root[p] += val_from_root;
    it_to_root[p] += val_to_root;
    if (l != r) {
      lz_from_root[p * 2] += val_from_root;
      lz_from_root[p * 2 + 1] += val_from_root;
      lz_to_root[p * 2] += val_to_root;
      lz_to_root[p * 2 + 1] += val_to_root;
    }
    return;
  }
  int m = (l + r) / 2;
  update(p * 2, l, m, ql, qr, val_from_root, val_to_root);
  update(p * 2 + 1, m + 1, r, ql, qr, val_from_root, val_to_root);
  it_from_root[p] = min(it_from_root[p * 2], it_from_root[p * 2 + 1]);
  it_to_root[p] = min(it_to_root[p * 2], it_to_root[p * 2 + 1]);
}
long long get(int p, int l, int r, int ql, int qr, int type) {
  lazy(p, l, r);
  if (ql > r || l > qr) return 1e15;
  if (ql <= l && r <= qr) {
    if (!type)
      return it_from_root[p];
    else
      return it_to_root[p];
  }
  int m = (l + r) / 2;
  return min(get(p * 2, l, m, ql, qr, type),
             get(p * 2 + 1, m + 1, r, ql, qr, type));
}
void dfs(int u, long long dis) {
  l[u] = r[u] = ++timer;
  update(1, 1, n, l[u], l[u], dis, dis + edge_to_root[u]);
  for (auto t : a[u]) {
    int v = t.first, c = t.second;
    if (v == p[u]) continue;
    p[v] = u;
    dfs(v, dis + c);
    r[u] = max(r[u], r[v]);
  }
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 1; i < n; i++) {
    int u, v, c;
    cin >> u >> v >> c;
    edge[i] = make_tuple(u, v, c);
    a[u].push_back(make_pair(v, c));
    a[v].push_back(make_pair(u, c));
  }
  for (int i = n; i <= 2 * n - 2; i++) {
    int u, v, c;
    cin >> u >> v >> c;
    edge[i] = make_tuple(u, v, c);
    edge_to_root[u] = c;
  }
  dfs(1, 0);
  for (int i = 1; i <= q; i++) {
    int type, id, c, u, v, old_c;
    cin >> type;
    if (type == 1) {
      cin >> id >> c;
      tie(u, v, old_c) = edge[id];
      if (id < n) {
        if (u == p[v]) swap(u, v);
        update(1, 1, n, l[u], r[u], c - old_c, c - old_c);
      } else
        update(1, 1, n, l[u], l[u], 0, c - old_c);
      get<2>(edge[id]) = c;
    } else {
      cin >> u >> v;
      if (l[u] <= l[v] && r[v] <= r[u])
        cout << get(1, 1, n, l[v], l[v], 0) - get(1, 1, n, l[u], l[u], 0)
             << '\n';
      else
        cout << get(1, 1, n, l[u], r[u], 1) - get(1, 1, n, l[u], l[u], 0) +
                    get(1, 1, n, l[v], l[v], 0)
             << '\n';
    }
  }
}