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 + 1;
const long long INF = 1e18;
int n, Q;
int idx, dfn[N], mx[N], to[N];
int a[N << 1], b[N << 1];
long long c[N << 1], dtree[N], dgraph[N];
vector<pair<int, long long> > E[N];
struct lwh {
  int ls, rs;
  long long w, tag;
} sgt[N << 2];
inline int rint() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
inline long long rll() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
inline void dfs(int x, int fa) {
  dfn[x] = ++idx;
  to[dfn[x]] = x;
  int lim = E[x].size();
  for (int i = 0; i < lim; i++) {
    int y = E[x][i].first;
    if (y == fa) continue;
    long long z = E[x][i].second;
    dtree[y] = dtree[x] + z;
    dfs(y, x);
  }
  mx[x] = idx;
}
inline void pushup(int p) { sgt[p].w = min(sgt[p << 1].w, sgt[p << 1 | 1].w); }
inline void pushdown(int p) {
  if (!sgt[p].tag) return;
  sgt[p << 1].w += sgt[p].tag, sgt[p << 1].tag += sgt[p].tag;
  sgt[p << 1 | 1].w += sgt[p].tag, sgt[p << 1 | 1].tag += sgt[p].tag;
  sgt[p].tag = 0;
}
inline void Build(int p, int l, int r) {
  sgt[p].ls = l, sgt[p].rs = r;
  if (l == r) {
    int pos = to[l];
    sgt[p].w = dtree[pos] + dgraph[pos];
    return;
  }
  int mid = (l + r) >> 1;
  Build(p << 1, l, mid);
  Build(p << 1 | 1, mid + 1, r);
  pushup(p);
}
inline void update(int p, int l, int r, long long k) {
  if (l > sgt[p].rs || r < sgt[p].ls) return;
  if (l <= sgt[p].ls && r >= sgt[p].rs) {
    sgt[p].w += k;
    sgt[p].tag += k;
    return;
  }
  pushdown(p);
  update(p << 1, l, r, k);
  update(p << 1 | 1, l, r, k);
  pushup(p);
}
inline long long query(int p, int l, int r) {
  if (l > sgt[p].rs || r < sgt[p].ls) return INF;
  if (l <= sgt[p].ls && r >= sgt[p].rs) return sgt[p].w;
  pushdown(p);
  long long lw = query(p << 1, l, r), rw = query(p << 1 | 1, l, r);
  pushup(p);
  return min(lw, rw);
}
inline void prework_on_tree() {
  for (int j = 0, i = 1; i < n; i++) {
    j++;
    int x = rint(), y = rint();
    long long z = rll();
    E[x].push_back(make_pair(y, z));
    a[j] = x, b[j] = y, c[j] = z;
  }
  dfs(1, 0);
}
inline void prework_on_graph() {
  for (int j = n - 1, i = 1; i < n; i++) {
    j++;
    int x = rint(), y = rint();
    long long z = rll();
    a[j] = x, b[j] = y, c[j] = z;
    dgraph[x] = z;
  }
  Build(1, 1, n);
}
int main() {
  n = rint(), Q = rint();
  prework_on_tree();
  prework_on_graph();
  while (Q--) {
    int tmp = rint();
    if (tmp == 1) {
      int id = rint();
      long long k = rint();
      if (id < n) {
        update(1, dfn[b[id]], mx[b[id]], k - c[id]);
        c[id] = k;
      } else {
        update(1, dfn[a[id]], dfn[a[id]], k - c[id]);
        c[id] = k;
        dgraph[a[id]] = k;
      }
    } else {
      int u = rint(), v = rint();
      if (dfn[v] >= dfn[u] && dfn[v] <= mx[u]) {
        long long du = query(1, dfn[u], dfn[u]) - dgraph[u],
                  dv = query(1, dfn[v], dfn[v]) - dgraph[v];
        printf("%lld\n", dv - du);
      } else {
        long long ans = query(1, dfn[u], mx[u]);
        long long du = query(1, dfn[u], dfn[u]) - dgraph[u],
                  dv = query(1, dfn[v], dfn[v]) - dgraph[v];
        ans -= du, ans += dv;
        printf("%lld\n", ans);
      }
    }
  }
  return 0;
}