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 long long int N = 4e5 + 2, inf = 1e12 + 7;
vector<pair<int, int> > vec[N];
pair<int, pair<int, int> > edg[N];
long long int n, dis[N], dis2[N], st[N], ed[N];
long long int T[2 * N], lzy[2 * N], tim;
void dfs(int u, int pr) {
  st[u] = ++tim;
  for (auto x : vec[u]) {
    int v = x.first, w = x.second;
    if (v == pr) continue;
    dis[v] = dis[u] + w;
    dfs(v, u);
  }
  ed[u] = tim;
}
void push_down(int nod, int lo, int hi) {
  T[nod] += lzy[nod];
  if (lo != hi) {
    lzy[2 * nod] += lzy[nod];
    lzy[2 * nod + 1] += lzy[nod];
  }
  lzy[nod] = 0;
}
void updt(int nod, int lo, int hi, int l, int r, int vl) {
  if (lzy[nod]) push_down(nod, lo, hi);
  if (hi < l || lo > r || lo > hi) return;
  if (lo >= l && hi <= r) {
    T[nod] += vl;
    if (lo != hi) {
      lzy[2 * nod] += vl;
      lzy[2 * nod + 1] += vl;
    }
    return;
  }
  int md = (lo + hi) >> 1;
  updt(2 * nod, lo, md, l, r, vl);
  updt(2 * nod + 1, md + 1, hi, l, r, vl);
  T[nod] = min(T[2 * nod], T[2 * nod + 1]);
}
long long int qry(int nod, int lo, int hi, int l, int r) {
  if (lzy[nod]) push_down(nod, lo, hi);
  if (hi < l || lo > r || lo > hi) return inf;
  if (lo >= l && hi <= r) return T[nod];
  int md = (lo + hi) >> 1;
  return min(qry(2 * nod, lo, md, l, r), qry(2 * nod + 1, md + 1, hi, l, r));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tc = 1, t = 0;
  while (tc--) {
    int q;
    cin >> n >> q;
    for (int i = 1; i < n; i++) {
      int u, v, c;
      cin >> u >> v >> c;
      vec[u].push_back(make_pair(v, c));
      edg[i] = make_pair(u, make_pair(v, c));
    }
    for (int i = n; i < n + n - 1; i++) {
      int u, v, c;
      cin >> u >> v >> c;
      edg[i] = make_pair(u, make_pair(v, c));
      dis2[u] = c;
    }
    dfs(1, -1);
    for (int i = 1; i <= n; i++) updt(1, 1, n, st[i], st[i], dis[i] + dis2[i]);
    while (q--) {
      int typ;
      cin >> typ;
      if (typ == 1) {
        int id, w;
        cin >> id >> w;
        int u = edg[id].first, v = edg[id].second.first;
        int c = edg[id].second.second;
        if (id < n) {
          updt(1, 1, n, st[v], ed[v], -c);
          updt(1, 1, n, st[v], ed[v], w);
          edg[id].second.second = w;
        } else {
          updt(1, 1, n, st[u], st[u], -c);
          updt(1, 1, n, st[u], st[u], w);
          edg[id].second.second = w;
          dis2[u] = w;
        }
      } else {
        int u, v;
        cin >> u >> v;
        if (st[v] >= st[u] && ed[u] >= ed[v]) {
          cout << qry(1, 1, n, st[v], st[v]) - qry(1, 1, n, st[u], st[u]) +
                      dis2[u] - dis2[v]
               << "\n";
        } else {
          long long int rs = qry(1, 1, n, st[u], ed[u]);
          rs -= qry(1, 1, n, st[u], st[u]);
          rs += qry(1, 1, n, st[v], st[v]);
          rs += dis2[u] - dis2[v];
          cout << rs << "\n";
        }
      }
    }
  }
  return 0;
}