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;
int n;
vector<int> e[200005];
int u[400005], v[400005], w[400005];
int l[200005], r[200005], t;
long long d0[200005], tmh[200005];
void dfs(int x, long long d) {
  l[x] = ++t;
  d0[x] = d;
  for (int eid : e[x]) {
    dfs(v[eid], d + w[eid]);
  }
  r[x] = t;
}
const int MAXN = 262144;
struct tree {
  long long mn[2 * MAXN], lazy[2 * MAXN];
  void init() {
    for (int i = MAXN - 1; i >= 1; i--) {
      mn[i] = min(mn[2 * i], mn[2 * i + 1]);
    }
  }
  void push(int x) {
    if (x < MAXN) {
      lazy[2 * x] += lazy[x];
      lazy[2 * x + 1] += lazy[x];
    }
    mn[x] += lazy[x];
    lazy[x] = 0;
  }
  void add(int l, int r, long long val, int node = 1, int nl = 1,
           int nr = MAXN) {
    push(node);
    if (r < nl || nr < l) {
      return;
    }
    if (l <= nl && nr <= r) {
      lazy[node] += val;
      push(node);
      return;
    }
    int nm = (nl + nr) >> 1;
    add(l, r, val, 2 * node, nl, nm);
    add(l, r, val, 2 * node + 1, nm + 1, nr);
    mn[node] = min(mn[2 * node], mn[2 * node + 1]);
  }
  long long get(int l, int r, int node = 1, int nl = 1, int nr = MAXN) {
    push(node);
    if (r < nl || nr < l) {
      return 1e18;
    }
    if (l <= nl && nr <= r) {
      return mn[node];
    }
    int nm = (nl + nr) >> 1;
    return min(get(l, r, 2 * node, nl, nm),
               get(l, r, 2 * node + 1, nm + 1, nr));
  }
};
tree drvo_dubina, drvo_takemehome;
void update_w(int eid, int nuw) {
  int delta = nuw - w[eid];
  w[eid] += delta;
  if (eid < n) {
    drvo_dubina.add(l[v[eid]], r[v[eid]], delta);
    drvo_takemehome.add(l[v[eid]], r[v[eid]], delta);
  } else {
    drvo_takemehome.add(l[u[eid]], l[u[eid]], delta);
  }
}
long long find_len(int a, int b) {
  if (l[a] <= l[b] && r[b] <= r[a]) {
    return drvo_dubina.get(l[b], l[b]) - drvo_dubina.get(l[a], l[a]);
  } else {
    return drvo_takemehome.get(l[a], r[a]) - drvo_dubina.get(l[a], l[a]) +
           find_len(1, b);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  int q;
  cin >> n >> q;
  for (int i = 1; i <= 2 * n - 2; i++) {
    cin >> u[i] >> v[i] >> w[i];
    if (i < n) {
      e[u[i]].push_back(i);
    } else {
      tmh[u[i]] = w[i];
    }
  }
  dfs(1, 0);
  for (int i = 1; i <= n; i++) {
    drvo_dubina.add(l[i], l[i], d0[i]);
  }
  for (int i = 1; i <= n; i++) {
    drvo_takemehome.add(l[i], l[i], d0[i] + tmh[i]);
  }
  while (q--) {
    int t, x, y;
    cin >> t >> x >> y;
    if (t == 1) {
      update_w(x, y);
    } else {
      cout << find_len(x, y) << '\n';
    }
  }
}