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 pra[maxn], pab[maxn];
vector<int> gra[maxn];
int n, q;
long long svor[2 * maxn];
int pradzianr[maxn];
int pabaiganr[maxn];
int keisk[2 * maxn];
struct minimumai {
  const long long static inf = 200000000000ll;
  long long val = 0;
  long long lazy = 0;
  int fr, to;
  minimumai *left = NULL, *right = NULL;
  void push() {
    val += lazy;
    right->lazy += lazy;
    left->lazy += lazy;
    lazy = 0;
  }
  minimumai(int fr, int to) : fr(fr), to(to) {
    if (fr == to) return;
    int vid = (fr + to) / 2;
    left = new minimumai(fr, vid);
    right = new minimumai(vid + 1, to);
  }
  void update(int f, int t, long long l) {
    if (to < f or t < fr) return;
    if (f <= fr and to <= t) {
      lazy += l;
      return;
    }
    push();
    right->update(f, t, l);
    left->update(f, t, l);
    val = min(right->val + right->lazy, left->val + left->lazy);
  }
  long long get(int f, int t) {
    if (to < f or t < fr) return inf;
    if (f <= fr and to <= t) return val + lazy;
    assert(fr != to);
    push();
    long long ret = min(right->get(f, t), left->get(f, t));
    return ret;
  }
} * medis;
int timer = 0;
void dfs(int d, long long kel = 0) {
  pra[d] = ++timer;
  medis->update(timer, timer, kel + svor[pradzianr[d]]);
  for (auto &&x : gra[d]) dfs(x, kel + svor[pabaiganr[x]]);
  pab[d] = timer;
}
long long inline kelias(int x) {
  return medis->get(pra[x], pra[x]) - svor[pradzianr[x]];
};
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b >> svor[i];
    a--, b--;
    pabaiganr[b] = i;
    keisk[i] = b;
    gra[a].push_back(b);
  }
  for (int i = n - 1; i < 2 * n - 2; i++) {
    int b;
    cin >> keisk[i] >> b >> svor[i];
    keisk[i]--;
    pradzianr[keisk[i]] = i;
  }
  pradzianr[0] = 2 * n - 1;
  svor[2 * n - 1] = 0;
  medis = new minimumai(0, maxn);
  dfs(0);
  for (int i = 0; i < q; i++) {
    int kom, u, v;
    cin >> kom >> u >> v;
    u--;
    if (kom == 1) {
      if (u < n - 1)
        medis->update(pra[keisk[u]], pab[keisk[u]], v - svor[u]);
      else
        medis->update(pra[keisk[u]], pra[keisk[u]], v - svor[u]);
      svor[u] = v;
    } else {
      v--;
      long long ku = kelias(u), kv = kelias(v);
      if (pra[u] <= pra[v] and pab[v] <= pab[u])
        cout << kv - ku << "\n";
      else
        cout << medis->get(pra[u], pab[u]) - ku + kv << "\n";
    }
  }
  return 0;
}