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
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200000 + 2;
const long long inf = (long long)1000000 * 1000000000;
long long p[MAXN * 4];
long long minimum[MAXN * 4];
int n, q;
inline void maintain(int v) {
  minimum[v] = min(minimum[2 * v + 1], minimum[2 * v + 2]);
}
inline void push(int v) {
  if (p[v] == 0) return;
  p[2 * v + 1] += p[v];
  p[2 * v + 2] += p[v];
  minimum[2 * v + 1] += p[v];
  minimum[2 * v + 2] += p[v];
  maintain(v);
  p[v] = 0;
}
void update(int v, int tl, int tr, int l, int r, long long delta) {
  if (l > r) return;
  if (tl == l & tr == r) {
    minimum[v] += delta;
    p[v] += delta;
    return;
  }
  int tm = (tl + tr) / 2;
  push(v);
  update(2 * v + 1, tl, tm, l, min(r, tm), delta);
  update(2 * v + 2, tm + 1, tr, max(tm + 1, l), r, delta);
  maintain(v);
}
long long getmin(int v, int tl, int tr, int l, int r) {
  if (l > r) return inf;
  if (tl == l && tr == r) return minimum[v];
  int tm = (tl + tr) / 2;
  push(v);
  long long first = getmin(2 * v + 1, tl, tm, l, min(r, tm));
  long long second = getmin(2 * v + 2, tm + 1, tr, max(l, tm + 1), r);
  maintain(v);
  return min(first, second);
}
struct edge {
  int initial, terminal;
  long long weight;
};
long long toroot[MAXN];
vector<struct edge> edges;
vector<int> g[MAXN];
int timer = -1;
int tin[MAXN];
int tout[MAXN];
void read() {
  scanf("%d %d", &n, &q);
  for (int i = 0; i < 2 * (n - 1); i++) {
    int initial, terminal;
    long long weight;
    scanf("%d %d %lld", &initial, &terminal, &weight);
    struct edge tmp;
    tmp.initial = initial;
    tmp.terminal = terminal;
    tmp.weight = weight;
    edges.push_back(tmp);
    if (terminal != 1) g[initial].push_back(edges.size() - 1);
  }
}
void dfs(int v, int p) {
  tin[v] = ++timer;
  for (auto index : g[v]) {
    int to = edges[index].terminal;
    if (to == p) continue;
    dfs(to, v);
  }
  tout[v] = timer;
}
inline bool upper(int a, int b) {
  return tin[a] <= tin[b] && tout[b] <= tout[a];
}
void filltree() {
  for (auto it : edges) {
    if (it.terminal == 1) {
      update(0, 0, n - 1, tin[it.initial], tin[it.initial], it.weight);
      toroot[it.initial] = it.weight;
    } else
      update(0, 0, n - 1, tin[it.terminal], tout[it.terminal], it.weight);
  }
}
void updateedge(int index, long long newval) {
  long long delta = newval - edges[index].weight;
  edges[index].weight = newval;
  int initial = edges[index].initial, terminal = edges[index].terminal;
  if (terminal == 1) {
    update(0, 0, n - 1, tin[initial], tin[initial], delta);
    toroot[initial] = newval;
  } else
    update(0, 0, n - 1, tin[terminal], tout[terminal], delta);
}
inline long long shortestwaytoroot(int v) {
  long long answer = getmin(0, 0, n - 1, tin[v], tout[v]);
  answer -= (getmin(0, 0, n - 1, tin[v], tin[v]) - toroot[v]);
  return answer;
}
inline long long shortestwayfromroot(int v) {
  return getmin(0, 0, n - 1, tin[v], tin[v]) - toroot[v];
}
long long decide(int a, int b) {
  if (upper(a, b)) return shortestwayfromroot(b) - shortestwayfromroot(a);
  return shortestwaytoroot(a) + shortestwayfromroot(b);
}
int main() {
  read();
  dfs(1, 1);
  filltree();
  for (int i = 0; i < q; i++) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int index;
      long long weight;
      scanf("%d %lld", &index, &weight);
      updateedge(index - 1, weight);
    }
    if (type == 2) {
      int u, v;
      scanf("%d %d", &u, &v);
      printf("%lld\n", decide(u, v));
    }
  }
  return 0;
}
