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;
int n, q, u[404000], v[404000], w[404000], bak[404000] = {0};
struct R {
  int to, val, nex;
} r[202000];
int hea[202000] = {0}, cnt = 1;
int df1[202000], df2[202000], dft = 0;
long long an, ans;
namespace D {
long long d[202000];
void RE() { memset(d, 0, sizeof d); }
void add(int x, int v) {
  for (; x <= n; x += x & (-x)) d[x] += v;
}
void A(int st, int en, int val) {
  add(st, val);
  add(en + 1, -val);
}
long long Q(int x) {
  long long t = 0;
  for (; x > 0; x -= x & -x) t += d[x];
  return t;
}
}  // namespace D
namespace tr {
long long mn[808000], laz[808000];
void U(int x) { mn[x] = min(mn[x << 1], mn[x << 1 | 1]); }
void D(int x) {
  if (!laz[x]) return;
  mn[x << 1] += laz[x];
  laz[x << 1] += laz[x];
  mn[x << 1 | 1] += laz[x];
  laz[x << 1 | 1] += laz[x];
  U(x);
  laz[x] = 0;
}
void B(int x, int l, int r) {
  laz[x] = 0;
  if (l == r) {
    mn[x] = D::Q(l) + bak[l];
    return;
  }
  int mid = (l + r) >> 1;
  B(x << 1, l, mid);
  B(x << 1 | 1, mid + 1, r);
  U(x);
}
void RE() { B(1, 1, n); }
long long que(int x, int l, int r, int st, int en) {
  if (st > en) return (1e16);
  if (l == st && r == en) return mn[x];
  D(x);
  int mid = (l + r) >> 1;
  return min(que(x << 1, l, mid, max(st, l), min(en, mid)),
             que(x << 1 | 1, mid + 1, r, max(st, mid + 1), min(en, r)));
}
long long Q(int st, int en) { return que(1, 1, n, st, en); }
void chg(int x, int l, int r, int st, int en, int val) {
  if (st > en) return;
  if (l == st && r == en) {
    mn[x] += val;
    laz[x] += val;
    return;
  }
  D(x);
  int mid = (l + r) >> 1;
  chg(x << 1, l, mid, max(st, l), min(en, mid), val);
  chg(x << 1 | 1, mid + 1, r, max(st, mid + 1), min(en, r), val);
  U(x);
}
void C(int st, int en, int val) { chg(1, 1, n, st, en, val); }
}  // namespace tr
void dfs(int x) {
  df1[x] = ++dft;
  int y;
  for (int i = hea[x]; i > 0; i = r[i].nex) {
    y = r[i].to;
    dfs(y);
  }
  df2[x] = dft;
}
int main() {
  D::RE();
  scanf("%d%d", &n, &q);
  int x, y, z;
  for (int i = 1; i < n; i++) {
    scanf("%d%d%d", &u[i], &v[i], &w[i]);
    r[++cnt] = (R){v[i], w[i], hea[u[i]]};
    hea[u[i]] = cnt;
  }
  dfs(1);
  for (int i = n; i <= n * 2 - 2; i++) {
    scanf("%d%d%d", &u[i], &v[i], &w[i]);
    bak[df1[u[i]]] = w[i];
  }
  for (int i = 1; i < n; i++) {
    D::A(df1[v[i]], df2[v[i]], w[i]);
  }
  tr::RE();
  for (int t; q--;) {
    scanf("%d%d%d", &t, &x, &y);
    if (t == 1) {
      if (x < n) {
        D::A(df1[v[x]], df2[v[x]], y - w[x]);
        tr::C(df1[v[x]], df2[v[x]], y - w[x]);
        w[x] = y;
      } else {
        tr::C(df1[u[x]], df1[u[x]], y - bak[df1[u[x]]]);
        bak[df1[u[x]]] = y;
      }
    } else {
      if (df1[x] <= df1[y] && df2[x] >= df2[y]) {
        ans = D::Q(df1[y]) - D::Q(df1[x]);
      } else {
        ans = tr::Q(df1[x], df2[x]) - D::Q(df1[x]) + D::Q(df1[y]);
      }
      printf("%I64d\n", ans);
    }
  }
}