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 = 200000;
int n, q, pra[MAXN], pab[MAXN], pabaiga[2 * MAXN], pabaiganr[2 * MAXN],
    pradzia[2 * MAXN], pradzianr[2 * MAXN], timer;
vector<int> g[MAXN];
long long c[2 * MAXN];
struct minimum {
  const long long INF = 2e11;
  long long val = 0, lazy = 0;
  int from, to;
  minimum *left = NULL, *right = NULL;
  void push() {
    val += lazy;
    if (right != NULL) right->lazy += lazy;
    if (left != NULL) left->lazy += lazy;
    lazy = 0;
  }
  void upd() {
    assert(right != NULL && left != NULL);
    val = min(right->val + right->lazy, left->val + left->lazy);
  }
  minimum(int from, int to) : from(from), to(to) {
    if (from == to) return;
    int vid = (from + to) / 2;
    left = new minimum(from, vid);
    right = new minimum(vid + 1, to);
  }
  void update(int f, int t, long long l) {
    if (to < f || t < from) return;
    if (f <= from && to <= t) {
      lazy += l;
      return;
    }
    push();
    right->update(f, t, l);
    left->update(f, t, l);
    upd();
  }
  long long get(int f, int t) {
    if (to < f || t < from) return INF;
    if (f <= from && to <= t) return val + lazy;
    assert(from != to);
    push();
    long long ret = min(right->get(f, t), left->get(f, t));
    return ret;
  }
} * medis;
void dfs(int v, long long d = 0) {
  pra[v] = ++timer;
  medis->update(timer, timer, d + c[pradzianr[v]]);
  for (auto &&u : g[v]) dfs(u, d + c[pabaiganr[u]]);
  pab[v] = timer;
}
long long kelias(int v) { return medis->get(pra[v], pra[v]) - c[pradzianr[v]]; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int type, a, b, u, v, i;
  cin >> n >> q;
  for (i = 0; i < n - 1; i++) {
    cin >> a >> b >> c[i];
    a--, b--;
    pabaiganr[b] = i;
    pabaiga[i] = b;
    g[a].push_back(b);
  }
  for (i = n - 1; i < 2 * n - 2; i++) {
    cin >> pradzia[i] >> b >> c[i];
    pradzia[i]--;
    pradzianr[pradzia[i]] = i;
  }
  pradzianr[0] = 2 * n - 1;
  c[2 * n - 1] = 0;
  medis = new minimum(0, 2 * MAXN + 2);
  dfs(0);
  for (i = 0; i < q; i++) {
    cin >> type >> u >> v;
    u--;
    if (type == 1) {
      if (u < n - 1)
        medis->update(pra[pabaiga[u]], pab[pabaiga[u]], v - c[u]);
      else
        medis->update(pra[pradzia[u]], pra[pradzia[u]], v - c[u]);
      c[u] = v;
    } else {
      v--;
      long long ku = kelias(u), kv = kelias(v);
      if (pra[u] <= pra[v] && pab[v] <= pab[u])
        cout << kv - ku << endl;
      else
        cout << medis->get(pra[u], pab[u]) - ku + kv << endl;
    }
  }
  assert(q > 0 && q <= 200000);
  return 0;
}