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;
vector<pair<long long, long long> > v[400005];
long long bk[400005] = {0};
long long x[400005] = {0}, y[400005] = {0}, w[400005] = {0};
long long de[400005] = {0}, st[400005] = {0}, en[400005] = {0}, tym = 0;
void dfs(long long r, long long val = 0) {
  st[r] = ++tym;
  de[r] = val + bk[r];
  for (auto i : v[r]) {
    dfs(i.first, val + i.second);
  }
  en[r] = tym;
}
struct Seg {
  vector<long long> tree, lzy;
  vector<bool> check;
  Seg(long long n) {
    tree.resize(4 * n, 0);
    lzy.resize(4 * n, 0);
  }
  void update(long long id, long long le, long long ri) {
    tree[id] = min(tree[le], tree[ri]) + lzy[id];
  }
  void upd(long long id, long long s, long long e, long long l, long long r,
           long long x) {
    if (l > e || s > r) return;
    if (l <= s && e <= r) {
      lzy[id] += x;
      tree[id] += x;
      return;
    }
    long long mid = (s + e) / 2;
    upd(id + id, s, mid, l, r, x);
    upd(id + id + 1, mid + 1, e, l, r, x);
    update(id, id + id, id + id + 1);
  }
  long long Query(long long id, long long s, long long e, long long l,
                  long long r) {
    if (l > e || s > r) return LLONG_MAX;
    if (l <= s && e <= r) return tree[id];
    long long mid = (s + e) / 2;
    return min(Query(id + id, s, mid, l, r),
               Query(id + id + 1, mid + 1, e, l, r)) +
           lzy[id];
  }
} * SGT;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, q;
  cin >> n >> q;
  SGT = new Seg(n + 1);
  for (long long i = 1; i < (2 * n - 1); i++) {
    long long a, b, c;
    cin >> a >> b >> c;
    if (i < n) {
      v[a].push_back({b, c});
    } else {
      bk[a] = c;
    }
    x[i] = a;
    y[i] = b;
    w[i] = c;
  }
  dfs(1);
  for (long long i = 1; i <= n; i++) {
    SGT->upd(1, 1, n, st[i], st[i], de[i]);
  }
  for (long long i = 1; i <= q; i++) {
    long long ty;
    cin >> ty;
    if (ty == 1) {
      long long ind, ww;
      cin >> ind >> ww;
      if (ind >= n) {
        SGT->upd(1, 1, n, st[x[ind]], st[x[ind]], -w[ind]);
        w[ind] = ww;
        SGT->upd(1, 1, n, st[x[ind]], st[x[ind]], w[ind]);
        bk[x[ind]] = ww;
      } else {
        SGT->upd(1, 1, n, st[y[ind]], en[y[ind]], -w[ind]);
        w[ind] = ww;
        SGT->upd(1, 1, n, st[y[ind]], en[y[ind]], w[ind]);
      }
    } else {
      long long uu, vv;
      cin >> uu >> vv;
      if (st[uu] <= st[vv] && en[uu] >= en[vv]) {
        long long ans = SGT->Query(1, 1, n, st[vv], st[vv]) - bk[vv];
        long long sub = SGT->Query(1, 1, n, st[uu], st[uu]) - bk[uu];
        ans -= sub;
        cout << ans << endl;
      } else {
        long long ans = SGT->Query(1, 1, n, st[uu], en[uu]);
        long long sub = SGT->Query(1, 1, n, st[uu], st[uu]) - bk[uu];
        ans -= sub;
        long long add = SGT->Query(1, 1, n, st[vv], st[vv]) - bk[vv];
        ans += add;
        cout << ans << endl;
      }
    }
  }
  return 0;
}