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 MAX_N = 5e5 + 10, INF = 999999999;
int n, q, from[MAX_N], to[MAX_N], w[MAX_N], be[MAX_N], st[MAX_N], en[MAX_N],
    dis[MAX_N], seg[4 * MAX_N], lazy[4 * MAX_N], sdis[4 * MAX_N],
    ldis[4 * MAX_N];
vector<pair<int, int>> adj[MAX_N];
vector<int> sten;
void dfs(int v, int p) {
  sten.push_back(v);
  st[v] = sten.size() - 1;
  for (pair<int, int> u : adj[v]) {
    ;
    if (u.first != p) {
      dis[u.first] = dis[v] + u.second;
      dfs(u.first, v);
    }
  }
  en[v] = sten.size();
}
void build(int id, int b, int e) {
  if (e - b == 1) {
    seg[id] = dis[sten[b]] + be[sten[b]];
    return;
  }
  int mid = (b + e) / 2;
  build(id * 2, b, mid);
  build(id * 2 + 1, mid, e);
  seg[id] = min(seg[id * 2], seg[id * 2 + 1]);
}
void shift(int id, int b, int e) {
  seg[id * 2] += lazy[id];
  lazy[id * 2] += lazy[id];
  seg[id * 2 + 1] += lazy[id];
  lazy[id * 2 + 1] += lazy[id];
  lazy[id] = 0;
}
void upd(int id, int b, int e, int l, int r, int v) {
  if (b >= r || l >= e) {
    return;
  }
  if (l <= b && e <= r) {
    seg[id] += v;
    lazy[id] += v;
    return;
  }
  shift(id, b, e);
  int mid = (b + e) / 2;
  upd(id * 2, b, mid, l, r, v);
  upd(id * 2 + 1, mid, e, l, r, v);
  seg[id] = min(seg[id * 2], seg[id * 2 + 1]);
}
int get(int id, int b, int e, int l, int r) {
  if (b >= r || l >= e) {
    return INF;
  }
  if (l <= b && e <= r) {
    return seg[id];
  }
  shift(id, b, e);
  int mid = (b + e) / 2;
  return min(get(id * 2, b, mid, l, r), get(id * 2 + 1, mid, e, l, r));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n - 1; i++) {
    cin >> from[i] >> to[i] >> w[i];
    adj[from[i]].push_back({to[i], w[i]});
    adj[to[i]].push_back({from[i], w[i]});
  }
  for (int i = 1; i <= n - 1; i++) {
    cin >> from[n - 1 + i] >> to[n - 1 + i] >> w[n - 1 + i];
    be[from[n - 1 + i]] = w[n - 1 + i];
  }
  sten.push_back(0);
  dfs(1, 0);
  build(1, 1, n + 1);
  while (q--) {
    int a, b, c;
    cin >> a >> b >> c;
    if (a == 1) {
      int x = c - w[b];
      w[b] = c;
      if (b <= n - 1) {
        upd(1, 1, n + 1, max(st[from[b]], st[to[b]]),
            min(en[from[b]], en[to[b]]), x);
      } else {
        be[from[b]] = c;
        upd(1, 1, n + 1, st[from[b]], st[from[b]] + 1, x);
      }
    } else {
      if (b == c) {
        cout << 0 << "\n";
      } else if (st[b] < st[c] && en[b] >= en[c]) {
        cout << (get(1, 1, n + 1, st[c], st[c] + 1) - be[c]) -
                    (get(1, 1, n + 1, st[b], st[b] + 1) - be[b])
             << "\n";
      } else {
        cout << (get(1, 1, n + 1, st[c], st[c] + 1) - be[c]) +
                    get(1, 1, n + 1, st[b], en[b]) -
                    (get(1, 1, n + 1, st[b], st[b] + 1) - be[b])
             << "\n";
      }
    }
  }
}