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;
struct edge {
  int u, v, d;
} edges[maxn];
int n, q, back[maxn], front[maxn], dist[maxn], t, tin[maxn], tout[maxn];
vector<pair<int, int> > grafo[maxn];
long long tree[4 * maxn], lazy[4 * maxn];
void flush(int node, int l, int r) {
  if (!lazy[node]) return;
  tree[node] += lazy[node];
  if (l != r) {
    lazy[2 * node] += lazy[node];
    lazy[2 * node + 1] += lazy[node];
  }
  lazy[node] = 0;
}
void update(int node, int tl, int tr, int l, int r, long long v) {
  flush(node, tl, tr);
  if (tl > r or tr < l) return;
  if (tl >= l and tr <= r) {
    lazy[node] = v;
    flush(node, tl, tr);
    return;
  }
  int mid = (tl + tr) >> 1;
  update(2 * node, tl, mid, l, r, v);
  update(2 * node + 1, mid + 1, tr, l, r, v);
  tree[node] = min(tree[2 * node], tree[2 * node + 1]);
}
long long query(int node, int tl, int tr, int l, int r) {
  flush(node, tl, tr);
  if (tl > r or tr < l) return 10000000000000000;
  if (tl >= l and tr <= r) return tree[node];
  int mid = (tl + tr) >> 1;
  return min(query(2 * node, tl, mid, l, r),
             query(2 * node + 1, mid + 1, tr, l, r));
}
void dfs(int u, int p, long long d) {
  tin[u] = ++t;
  for (auto v : grafo[u]) {
    if (v.first == p) continue;
    dfs(v.first, u, d + v.second);
  }
  update(1, 1, n, tin[u], tin[u], d + dist[u]);
  tout[u] = t;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> q;
  for (int i = 1; i < n; i++) {
    int u, v, d;
    cin >> u >> v >> d;
    grafo[u].push_back({v, d});
    edges[i] = {u, v, d};
  }
  for (int i = n; i <= 2 * n - 2; i++) {
    int u, v, d;
    cin >> u >> v >> d;
    edges[i] = {u, v, d};
    dist[u] = d;
  }
  dfs(1, 0, 0);
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int id, d;
      cin >> id >> d;
      if (id < n) {
        int v = edges[id].v;
        update(1, 1, n, tin[v], tout[v], -edges[id].d);
        update(1, 1, n, tin[v], tout[v], d);
        edges[id].d = d;
      } else {
        int u = edges[id].u;
        update(1, 1, n, tin[u], tin[u], -dist[u]);
        update(1, 1, n, tin[u], tin[u], d);
        dist[u] = d;
      }
    } else {
      int u, v;
      cin >> u >> v;
      if (tin[v] >= tin[u] and tout[u] >= tout[v]) {
        cout << -query(1, 1, n, tin[u], tin[u]) +
                    query(1, 1, n, tin[v], tin[v]) + dist[u] - dist[v]
             << "\n";
      } else {
        cout << query(1, 1, n, tin[u], tout[u]) -
                    query(1, 1, n, tin[u], tin[u]) +
                    query(1, 1, n, tin[v], tin[v]) + dist[u] - dist[v]
             << "\n";
      }
    }
  }
  return 0;
}