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 = 233333;
int head[maxn], f[maxn], dis[maxn], dep[maxn], dis2[maxn];
int n, m, x, y, k;
int cnt;
struct edge {
  int fr, next, to, val;
} e[maxn << 1];
void add_edge(int x, int y, int v) {
  e[cnt].to = y;
  e[cnt].next = head[x];
  e[cnt].val = v;
  e[cnt].fr = x;
  head[x] = cnt++;
}
void update(int id, int k) {
  int v = e[id].fr;
  if (id < n) {
    dis2[e[id].to] = k;
  }
  e[id].val = k;
  int tmp = 0x3f3f3f3f;
  while (v != 1) {
    for (int i = head[v]; ~i; i = e[i].next) {
      int w = e[i].to;
      if (w == 1) tmp = e[i].val;
      if (dep[w] < dep[v]) continue;
      tmp = min(tmp, dis[w] + e[i].val);
    }
    if (tmp == dis[v])
      break;
    else {
      dis[v] = tmp;
      v = f[v];
    }
  }
}
void dfs(int u) {
  for (int i = head[u]; ~i; i = e[i].next) {
    int v = e[i].to;
    if (dep[v] > 0) continue;
    dep[v] = dep[u] + 1;
    dis2[v] = e[i].val;
    dfs(v);
    dis[u] = min(dis[u], dis[v] + e[i].val);
  }
}
int query(int x, int y) {
  int ans = 0;
  int v = y;
  if (dep[x] <= dep[y]) {
    while (dep[x] < dep[v]) {
      ans += dis2[v];
      v = f[v];
    }
    if (x == v) {
      return ans;
    }
  }
  while (v != 1) {
    ans += dis2[v];
    v = f[v];
  }
  return dis[x] + ans;
}
int main() {
  cnt = 1;
  cin >> n >> m;
  memset(dis, 0x3f, sizeof dis);
  memset(dep, -1, sizeof dep);
  memset(head, -1, sizeof head);
  dis[1] = 0;
  dep[1] = 1;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d%d", &x, &y, &k);
    f[y] = x;
    add_edge(x, y, k);
  }
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d%d", &x, &y, &k);
    dis[x] = k;
    add_edge(x, y, k);
  }
  dfs(1);
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &k, &x, &y);
    if (k == 1)
      update(x, y);
    else
      printf("%d\n", query(x, y));
  }
}