838_B. Diverging Directions

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

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


Example output:
0
1
4
8
100
132
10


Tags: data structures
dfs and similar
trees
Difficulty rating: 2100

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 200000 + 10;
const int INF = 0x3f3f3f3f;
int n, to[MAX_N], back[MAX_N], dep[MAX_N];
int head[MAX_N], tot = 0;
int from[MAX_N * 2], nxt[MAX_N * 2], p[MAX_N], v[MAX_N * 2], w[MAX_N * 2];
void add_edge(int a, int b, int c) {
  tot++;
  nxt[tot] = head[a];
  head[a] = tot;
  from[tot] = a;
  v[tot] = b;
  w[tot] = c;
}
void dfs(int u) {
  for (int i = head[u]; ~i; i = nxt[i]) {
    if (dep[v[i]] > 0) continue;
    dep[v[i]] = dep[u] + 1;
    to[v[i]] = w[i];
    dfs(v[i]);
    back[u] = min(back[u], w[i] + back[v[i]]);
  }
}
void update(int id, int W) {
  to[v[id]] = W;
  int u = from[id];
  w[id] = W;
  int res = INF;
  while (u != 1) {
    for (int i = head[u]; ~i; i = nxt[i]) {
      if (v[i] == 1) res = w[i];
      if (dep[v[i]] < dep[u]) continue;
      res = min(res, back[v[i]] + w[i]);
    }
    if (res == back[u])
      break;
    else {
      back[u] = res;
      u = p[u];
    }
  }
}
int query(int a, int b) {
  int ans = 0;
  if (dep[a] <= dep[b]) {
    while (dep[a] < dep[b]) {
      ans += to[b];
      b = p[b];
    }
    if (a == b) return ans;
  }
  while (b != 1) {
    ans += to[b];
    b = p[b];
  }
  return back[a] + ans;
}
int main() {
  int q;
  scanf("%d%d", &n, &q);
  int a, b, c;
  memset(head, -1, sizeof(head));
  memset(p, -1, sizeof(p));
  memset(back, INF, sizeof(back));
  back[1] = 0;
  dep[1] = 1;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d%d", &a, &b, &c);
    p[b] = a;
    add_edge(a, b, c);
  }
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d%d", &a, &b, &c);
    back[a] = c;
    add_edge(a, b, c);
  }
  dfs(1);
  while (q--) {
    scanf("%d%d%d", &a, &b, &c);
    if (a == 1) {
      update(b, c);
    } else
      printf("%d\n", query(b, c));
  }
  return 0;
}