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 INF = 2e9;
const int maxn = 200000 + 100;
vector<pair<int, int> > g[maxn];
int too[maxn];
int arr[maxn];
pair<int, int> sf[maxn];
int seg[4 * maxn];
int sfc;
int lazy[4 * maxn];
void dfs(int v, int d = 0, int p = 0) {
  sf[v].first = sfc;
  arr[sfc] = d + too[v];
  sfc++;
  for (pair<int, int> i : g[v])
    if (i.first != p) dfs(i.first, d + i.second, v);
  sf[v].second = sfc;
}
void make(int b, int e, int id) {
  if (e - b == 1) {
    seg[id] = arr[b];
    return;
  }
  int m = (b + e) / 2;
  make(b, m, id * 2);
  make(m, e, (id * 2) + 1);
  seg[id] = min(seg[id * 2], seg[(id * 2) + 1]);
}
void shift(int id) {
  int s = lazy[id];
  seg[id * 2] += s;
  seg[(id * 2) + 1] += s;
  lazy[id * 2] += s;
  lazy[(id * 2) + 1] += s;
  lazy[id] = 0;
}
void upd(int b, int e, int l, int r, int val, int id) {
  if (r <= b || e <= l) return;
  if (l <= b && e <= r) {
    seg[id] += val;
    lazy[id] += val;
    return;
  }
  shift(id);
  int m = (b + e) / 2;
  upd(b, m, l, r, val, id * 2);
  upd(m, e, l, r, val, (id * 2) + 1);
  seg[id] = min(seg[id * 2], seg[(id * 2) + 1]);
}
int get(int b, int e, int l, int r, int id) {
  if (r <= b || e <= l) return INF;
  if (l <= b && e <= r) return seg[id];
  shift(id);
  int m = (b + e) / 2;
  return min(get(b, m, l, r, id * 2), get(m, e, l, r, (id * 2) + 1));
}
pair<int, int> edge[maxn];
pair<int, int> edge2[maxn];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n - 1; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    g[u].push_back({v, w});
    edge[i + 1] = {v, w};
  }
  for (int i = 0; i < n - 1; i++) {
    int u, w;
    cin >> u >> w >> w;
    too[u] = w;
    edge2[i + 1] = {u, w};
  }
  dfs(1);
  make(0, n, 1);
  while (q--) {
    int a, b, c;
    cin >> a >> b >> c;
    if (a == 1) {
      if (b < n) {
        upd(0, n, sf[edge[b].first].first, sf[edge[b].first].second,
            c - edge[b].second, 1);
        edge[b].second = c;
      } else {
        b -= (n - 1);
        int s = sf[edge2[b].first].first;
        upd(0, n, s, s + 1, c - edge2[b].second, 1);
        edge2[b].second = c;
        too[edge2[b].first] = c;
      }
      continue;
    }
    int r = get(0, n, sf[b].first, sf[b].first + 1, 1);
    r -= too[b];
    int gh = get(0, n, sf[c].first, sf[c].first + 1, 1);
    gh -= too[c];
    if (sf[b].first <= sf[c].first && sf[b].second > sf[c].first) {
      cout << abs(r - gh) << endl;
      continue;
    }
    int kl = get(0, n, sf[b].first, sf[b].second, 1);
    cout << kl - r + gh << endl;
  }
}