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 = 2e5 + 10;
const long long INF = 1e17;
int n, m;
int in[maxn], out[maxn];
vector<int> vec[maxn];
int from[maxn * 2], to[maxn * 2];
long long c[maxn * 2];
struct Node {
  int l, r;
  long long val, add;
  int mid() { return (l + r) / 2; }
} node[maxn << 2];
struct Node2 {
  int l, r;
  long long val, add;
  int mid() { return (l + r) / 2; }
} node2[maxn << 2];
int cnt;
void dfs(int u) {
  in[u] = ++cnt;
  int v;
  for (int i = 0; i < vec[u].size(); i++) {
    v = vec[u][i];
    dfs(v);
  }
  out[u] = cnt;
}
void pushdown(int rt) {
  if (node[rt].add == 0) return;
  node[rt << 1].add += node[rt].add;
  node[rt << 1 | 1].add += node[rt].add;
  node[rt << 1].val += node[rt].add;
  node[rt << 1 | 1].val += node[rt].add;
  node[rt].add = 0;
}
void pushdown2(int rt) {
  if (node2[rt].add == 0) return;
  node2[rt << 1].add += node2[rt].add;
  node2[rt << 1 | 1].add += node2[rt].add;
  node2[rt << 1].val += node2[rt].add;
  node2[rt << 1 | 1].val += node2[rt].add;
  node2[rt].add = 0;
}
void build(int rt, int l, int r) {
  node[rt].l = l;
  node[rt].r = r;
  node[rt].add = 0;
  node[rt].val = 0;
  if (l == r) return;
  int mid = node[rt].mid();
  build(rt << 1, l, mid);
  build(rt << 1 | 1, mid + 1, r);
}
void build2(int rt, int l, int r) {
  node2[rt].l = l;
  node2[rt].r = r;
  node2[rt].add = 0;
  node2[rt].val = 0;
  if (l == r) return;
  int mid = node2[rt].mid();
  build2(rt << 1, l, mid);
  build2(rt << 1 | 1, mid + 1, r);
}
void update(int rt, int l, int r, long long xx) {
  if (node[rt].l == l && node[rt].r == r) {
    node[rt].val += xx;
    node[rt].add += xx;
    return;
  }
  pushdown(rt);
  int mid = node[rt].mid();
  if (r <= mid)
    update(rt << 1, l, r, xx);
  else if (l > mid)
    update(rt << 1 | 1, l, r, xx);
  else {
    update(rt << 1, l, mid, xx);
    update(rt << 1 | 1, mid + 1, r, xx);
  }
  node[rt].val = min(node[rt << 1].val, node[rt << 1 | 1].val);
}
void update2(int rt, int l, int r, long long xx) {
  if (node2[rt].l == l && node2[rt].r == r) {
    node2[rt].val += xx;
    node2[rt].add += xx;
    return;
  }
  pushdown2(rt);
  int mid = node2[rt].mid();
  if (r <= mid)
    update2(rt << 1, l, r, xx);
  else if (l > mid)
    update2(rt << 1 | 1, l, r, xx);
  else {
    update2(rt << 1, l, mid, xx);
    update2(rt << 1 | 1, mid + 1, r, xx);
  }
  node2[rt].val = min(node2[rt << 1].val, node2[rt << 1 | 1].val);
}
long long Min;
void query(int rt, int l, int r) {
  if (node[rt].l == l && node[rt].r == r) {
    Min = min(Min, node[rt].val);
    return;
  }
  pushdown(rt);
  int mid = node[rt].mid();
  if (r <= mid)
    query(rt << 1, l, r);
  else if (l > mid)
    query(rt << 1 | 1, l, r);
  else {
    query(rt << 1, l, mid);
    query(rt << 1 | 1, mid + 1, r);
  }
}
void query2(int rt, int l, int r) {
  if (node2[rt].l == l && node2[rt].r == r) {
    Min = min(Min, node2[rt].val);
    return;
  }
  pushdown2(rt);
  int mid = node2[rt].mid();
  if (r <= mid)
    query2(rt << 1, l, r);
  else if (l > mid)
    query2(rt << 1 | 1, l, r);
  else {
    query2(rt << 1, l, mid);
    query2(rt << 1 | 1, mid + 1, r);
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i < 2 * n - 1; i++)
    scanf("%d%d%lld", &from[i], &to[i], &c[i]);
  build(1, 1, n);
  build2(1, 1, n);
  for (int i = 1; i < n; i++) vec[from[i]].push_back(to[i]);
  dfs(1);
  int u, v;
  for (int i = 1; i < n; i++) {
    v = to[i];
    update(1, in[v], out[v], c[i]);
    update2(1, in[v], out[v], c[i]);
  }
  for (int i = n; i < 2 * n - 1; i++) {
    u = from[i];
    update2(1, in[u], in[u], c[i]);
  }
  int op;
  long long temp, temp2;
  int i;
  while (m--) {
    scanf("%d%d%d", &op, &u, &v);
    if (op == 1) {
      if (u < n) {
        i = u;
        u = to[u];
        update(1, in[u], out[u], (long long)v - c[i]);
        update2(1, in[u], out[u], (long long)v - c[i]);
        c[i] = v;
      } else {
        i = u;
        u = from[u];
        update2(1, in[u], in[u], (long long)v - c[i]);
        c[i] = v;
      }
    } else {
      Min = INF;
      if (in[u] <= in[v] && out[u] >= out[v]) {
        query(1, in[u], in[u]);
        temp = Min;
        Min = INF;
        query(1, in[v], in[v]);
        printf("%lld\n", Min - temp);
      } else {
        query2(1, in[u], out[u]);
        temp = Min;
        Min = INF;
        query(1, in[u], in[u]);
        temp2 = Min;
        Min = INF;
        query(1, in[v], in[v]);
        printf("%lld\n", Min + temp - temp2);
      }
    }
  }
  return 0;
}