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>
#pragma comment(linker, "/STACK:16000000")
using namespace std;
const int Maxn = 200005;
const int Maxm = 2 * Maxn;
const int Maxs = 1048576;
int n, q;
int a[Maxm], b[Maxm], c[Maxm];
vector<int> neigh[Maxn];
int my[Maxn];
int cur, lef[Maxn], rig[Maxn];
long long BIT[Maxn];
long long mn[Maxs], flag[Maxs];
void downOn(int v, long long fl) {
  mn[v] += fl;
  flag[v] += fl;
}
void Down(int v) {
  if (flag[v]) {
    downOn(2 * v, flag[v]);
    downOn(2 * v + 1, flag[v]);
    flag[v] = 0;
  }
}
void Union(int v) { mn[v] = min(mn[2 * v], mn[2 * v + 1]); }
void Update(int v, int l, int r, int a, int b, long long val) {
  if (l == a && r == b)
    downOn(v, val);
  else {
    Down(v);
    int m = l + r >> 1;
    if (a <= m) Update(2 * v, l, m, a, min(m, b), val);
    if (m + 1 <= b) Update(2 * v + 1, m + 1, r, max(m + 1, a), b, val);
    Union(v);
  }
}
void Change(int v, int l, int r, int x, long long val) {
  if (l == r) {
    mn[v] = val;
    flag[v] = 0;
  } else {
    Down(v);
    int m = l + r >> 1;
    if (x <= m)
      Change(2 * v, l, m, x, val);
    else
      Change(2 * v + 1, m + 1, r, x, val);
    Union(v);
  }
}
long long Get(int v, int l, int r, int a, int b) {
  if (l == a && r == b)
    return mn[v];
  else {
    Down(v);
    int m = l + r >> 1;
    if (b <= m) return Get(2 * v, l, m, a, b);
    if (m + 1 <= a) return Get(2 * v + 1, m + 1, r, a, b);
    return min(Get(2 * v, l, m, a, m), Get(2 * v + 1, m + 1, r, m + 1, b));
  }
}
void Traverse(int v) {
  lef[v] = ++cur;
  for (int i = 0; i < neigh[v].size(); i++) Traverse(neigh[v][i]);
  rig[v] = cur;
}
void Add(int x, long long val) {
  for (int i = x; i <= n; i += i & -i) BIT[i] += val;
}
long long Get(int x) {
  long long res = 0;
  for (int i = x; i > 0; i -= i & -i) res += BIT[i];
  return res;
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= 2 * n - 2; i++) {
    scanf("%d %d %d", &a[i], &b[i], &c[i]);
    if (i <= n - 1)
      neigh[a[i]].push_back(b[i]);
    else
      my[a[i]] = c[i];
  }
  Traverse(1);
  for (int i = 1; i <= n - 1; i++) {
    int v = b[i];
    Add(lef[v], c[i]);
    Add(rig[v] + 1, -c[i]);
  }
  for (int i = n; i <= 2 * n - 2; i++)
    Change(1, 1, n, lef[a[i]], Get(lef[a[i]]) + my[a[i]]);
  while (q--) {
    int typ, u, v;
    scanf("%d %d %d", &typ, &u, &v);
    if (typ == 1)
      if (u <= n - 1) {
        int mod = v - c[u];
        c[u] = v;
        Add(lef[b[u]], mod);
        Add(rig[b[u]] + 1, -mod);
        Update(1, 1, n, lef[b[u]], rig[b[u]], mod);
      } else {
        my[a[u]] = v;
        Change(1, 1, n, lef[a[u]], Get(lef[a[u]]) + my[a[u]]);
      }
    else if (lef[u] <= lef[v] && lef[v] <= rig[u])
      printf("%I64d\n", Get(lef[v]) - Get(lef[u]));
    else
      printf("%I64d\n",
             Get(1, 1, n, lef[u], rig[u]) - Get(lef[u]) + Get(lef[v]));
  }
  return 0;
}