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 M = (int)2e5 + 5, INF = (int)1e9 + 7;
template <class T>
void Rd(T &res) {
  res = 0;
  static char p;
  while (p = getchar(), p < '0')
    ;
  do {
    res = (res * 10) + (p ^ 48);
  } while (p = getchar(), p >= '0');
}
template <class T>
void Ps(T x) {
  if (x == 0) return;
  Ps(x / 10);
  putchar(x % 10 ^ 48);
}
template <class T>
void Pt(T x) {
  if (x == 0)
    putchar('0');
  else
    Ps(x);
  putchar('\n');
}
template <class T>
void Min(T &x, T y) {
  if (x > y) x = y;
}
struct W {
  int to, nx;
} Lis[M];
int Head[M], tot;
void Add(int x, int y) {
  Lis[++tot] = (W){y, Head[x]};
  Head[x] = tot;
}
int n, q, de[M], up_to_1[M], L[M], R[M], dfs_cnt, id[M * 2];
long long val[M * 4], che[M * 4], dep[M];
void ADD(int x, int v) {
  while (x <= n) {
    dep[x] += v;
    x += x & (-x);
  }
}
long long Sum(int x) {
  long long re = 0;
  while (x) {
    re += dep[x];
    x -= x & (-x);
  }
  return re;
}
void up(int p) { val[p] = min(val[p << 1], val[p << 1 | 1]); }
void down(int p) {
  if (che[p]) {
    che[p << 1] += che[p], che[p << 1 | 1] += che[p];
    val[p << 1] += che[p], val[p << 1 | 1] += che[p];
    che[p] = 0;
  }
}
void update(int L, int R, int p, int x, long long v) {
  if (L == R) {
    val[p] = v;
    return;
  }
  down(p);
  int mid = (L + R) >> 1;
  if (mid >= x)
    update(L, mid, p << 1, x, v);
  else
    update(mid + 1, R, p << 1 | 1, x, v);
  up(p);
}
void update_(int L, int R, int p, int l, int r, int v) {
  if (L == l && R == r) {
    val[p] += v;
    che[p] += v;
    return;
  }
  down(p);
  int mid = (L + R) >> 1;
  if (mid >= r)
    update_(L, mid, p << 1, l, r, v);
  else if (mid < l)
    update_(mid + 1, R, p << 1 | 1, l, r, v);
  else
    update_(L, mid, p << 1, l, mid, v),
        update_(mid + 1, R, p << 1 | 1, mid + 1, r, v);
  up(p);
}
long long query(int L, int R, int p, int l, int r) {
  if (L == l && R == r) return val[p];
  down(p);
  int mid = (L + R) >> 1;
  if (mid >= r)
    return query(L, mid, p << 1, l, r);
  else if (mid < l)
    return query(mid + 1, R, p << 1 | 1, l, r);
  else
    return min(query(L, mid, p << 1, l, mid),
               query(mid + 1, R, p << 1 | 1, mid + 1, r));
}
void dfs(int x) {
  L[x] = ++dfs_cnt;
  ADD(L[x], de[x]);
  for (int i = Head[x], to; i && (to = Lis[i].to, 1); i = Lis[i].nx) dfs(to);
  R[x] = dfs_cnt;
  ADD(R[x] + 1, -de[x]);
  update(1, n, 1, L[x], Sum(L[x]) + up_to_1[x]);
}
int main() {
  Rd(n), Rd(q);
  for (int i = 1, x, y; i < n; i++)
    Rd(x), Rd(y), Rd(de[y]), Add(x, y), id[i] = y;
  for (int i = 1, x, y; i < n; i++)
    Rd(x), Rd(y), Rd(up_to_1[x]), id[i + n - 1] = x;
  dfs(1);
  for (int i = 1, op, x, y; i <= q; i++) {
    Rd(op), Rd(x), Rd(y);
    if (op == 1) {
      if (x < n) {
        x = id[x];
        ADD(L[x], y - de[x]);
        ADD(R[x] + 1, de[x] - y);
        update_(1, n, 1, L[x], R[x], y - de[x]);
        de[x] = y;
      } else {
        x = id[x];
        up_to_1[x] = y;
        update(1, n, 1, L[x], up_to_1[x] + Sum(L[x]));
      }
    } else {
      long long ans = Sum(L[y]) + query(1, n, 1, L[x], R[x]) - Sum(L[x]);
      if (L[x] <= L[y] && R[x] >= R[y]) Min(ans, Sum(L[y]) - Sum(L[x]));
      Pt(ans);
    }
  }
  return 0;
}