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 = 2e5 + 10;
const long long INF = 1e18;
int n, q, in[MAXN], out[MAXN], cost[2 * MAXN], timer, u[2 * MAXN], v[2 * MAXN];
vector<pair<int, int>> g[MAXN];
long long t[MAXN * 4], w[MAXN * 4], d[MAXN];
void push(int v, int l, int r) {
  t[v] += w[v];
  if (l != r) {
    w[2 * v] += w[v];
    w[2 * v + 1] += w[v];
  }
  w[v] = 0;
}
void addAtRange(int v, int l, int r, int L, int R, long long x) {
  push(v, l, r);
  if (r < L || R < l || R < L) return;
  if (L <= l && r <= R) {
    w[v] += x;
    push(v, l, r);
    return;
  }
  int m = (l + r) / 2;
  addAtRange(2 * v, l, m, L, R, x);
  addAtRange(2 * v + 1, m + 1, r, L, R, x);
  t[v] = min(t[2 * v], t[2 * v + 1]);
}
long long getMin(int v, int l, int r, int L, int R) {
  push(v, l, r);
  if (r < L || R < l || R < L) return INF;
  if (L <= l && r <= R) return t[v];
  int m = (l + r) / 2;
  return min(getMin(2 * v, l, m, L, R), getMin(2 * v + 1, m + 1, r, L, R));
}
long long fenwickTree[MAXN];
void add(int i, int val) {
  for (; i < MAXN; i |= (i + 1)) {
    fenwickTree[i] += val;
  }
}
long long getD(int r) {
  long long total = d[r];
  for (; r >= 0; r = (r & (r + 1)) - 1) {
    total += fenwickTree[r];
  }
  return total;
}
void addAtRangeD(int l, int r, long long cost) {
  add(l, cost);
  add(r + 1, -cost);
}
void dfs(int v, int p = -1) {
  in[v] = ++timer;
  for (auto way : g[v]) {
    int to = way.first, cost = way.second;
    if (to != p) {
      d[timer + 1] = d[in[v]] + cost;
      dfs(to, v);
      addAtRange(1, 1, n, in[to], out[to], cost);
    }
  }
  out[v] = timer;
}
bool upper(int u, int v) { return in[u] <= in[v] && out[v] <= out[u]; }
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 0; i < 2 * n - 2; i++) {
    scanf("%d%d%d", &u[i], &v[i], &cost[i]);
    if (i < n - 1) {
      g[u[i]].push_back({v[i], cost[i]});
    }
  }
  dfs(1, 0);
  for (int i = n - 1; i < 2 * (n - 1); i++) {
    addAtRange(1, 1, n, in[u[i]], in[u[i]], cost[i]);
  }
  while (q--) {
    int type, a, b;
    scanf("%d%d%d", &type, &a, &b);
    if (type == 1) {
      a--;
      if (a < n - 1) {
        if (!upper(u[a], v[a])) swap(u[a], v[a]);
        addAtRangeD(in[v[a]], out[v[a]], -cost[a]);
        addAtRange(1, 1, n, in[v[a]], out[v[a]], -cost[a]);
        cost[a] = b;
        addAtRangeD(in[v[a]], out[v[a]], cost[a]);
        addAtRange(1, 1, n, in[v[a]], out[v[a]], cost[a]);
      } else {
        addAtRange(1, 1, n, in[u[a]], in[u[a]], -cost[a]);
        cost[a] = b;
        addAtRange(1, 1, n, in[u[a]], in[u[a]], cost[a]);
      }
    } else {
      long long dist = (upper(a, b) ? getD(in[b]) - getD(in[a]) : INF);
      dist =
          min(dist, getMin(1, 1, n, in[a], out[a]) - getD(in[a]) + getD(in[b]));
      cout << dist << "\n";
    }
  }
}