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 N = 2e5 + 5;
int n;
vector<pair<int, int> > adj[N];
int a[2 * N], b[2 * N], c[2 * N];
int sz[N], in[N], edgeToRoot[N], ind;
long long segTree[4 * N], lazy[4 * N];
void pushDown(int l, int r, int pos) {
  segTree[pos] += lazy[pos];
  if (l != r) {
    lazy[pos * 2] += lazy[pos];
    lazy[pos * 2 + 1] += lazy[pos];
  }
  lazy[pos] = 0;
}
void insert(int ind, long long v, int l = 0, int r = n - 1, int pos = 1) {
  if (ind > r || ind < l) return;
  if (l == r) {
    segTree[pos] = v;
    return;
  }
  int mid = (l + r) / 2;
  insert(ind, v, l, mid, pos * 2);
  insert(ind, v, mid + 1, r, pos * 2 + 1);
  segTree[pos] = min(segTree[pos * 2], segTree[pos * 2 + 1]);
}
void update(int s, int e, int v, int l = 0, int r = n - 1, int pos = 1) {
  pushDown(l, r, pos);
  if (l > e || r < s) return;
  if (l >= s && r <= e) {
    segTree[pos] += v;
    if (l != r) {
      lazy[pos * 2] += v;
      lazy[pos * 2 + 1] += v;
    }
    return;
  }
  int mid = (l + r) / 2;
  update(s, e, v, l, mid, pos * 2);
  update(s, e, v, mid + 1, r, pos * 2 + 1);
  segTree[pos] = min(segTree[pos * 2], segTree[pos * 2 + 1]);
}
long long query(int s, int e, int l = 0, int r = n - 1, int pos = 1) {
  pushDown(l, r, pos);
  if (l > e || r < s) return LLONG_MAX / 2;
  if (l >= s && r <= e) return segTree[pos];
  int mid = (l + r) / 2;
  return min(query(s, e, l, mid, pos * 2),
             query(s, e, mid + 1, r, pos * 2 + 1));
}
long long get(int node) {
  if (node == 1) return 0;
  return query(in[node], in[node]) - c[edgeToRoot[node]];
}
int dfs(int node, int par, long long dist) {
  sz[node] = 1;
  in[node] = ind;
  insert(ind++, dist + c[edgeToRoot[node]]);
  for (auto &c : adj[node])
    if (c.first != par) sz[node] += dfs(c.first, node, dist + c.second);
  return sz[node];
}
int main() {
  int q;
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d%d", a + i, b + i, c + i);
    adj[a[i]].push_back({b[i], c[i]});
  }
  for (int i = n - 1; i < 2 * n - 2; i++)
    scanf("%d%d%d", a + i, b + i, c + i), edgeToRoot[a[i]] = i;
  memset(segTree, 63, sizeof 63);
  dfs(1, 1, 0);
  int type, i, w, u, v;
  while (q--) {
    scanf("%d", &type);
    if (type == 1) {
      scanf("%d%d", &i, &w), i--;
      if (i < n - 1)
        update(in[b[i]], in[b[i]] + sz[b[i]] - 1, w - c[i]);
      else
        update(in[a[i]], in[a[i]], w - c[i]);
      c[i] = w;
    } else {
      scanf("%d%d", &u, &v);
      if (in[v] >= in[u] && in[v] < in[u] + sz[u])
        printf("%lld\n", get(v) - get(u));
      else
        printf("%lld\n", query(in[u], in[u] + sz[u] - 1) - get(u) + get(v));
    }
  }
}