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 MAXN = 200000 + 10;
int n, q, id;
int st[MAXN], edd[MAXN], head[MAXN], path[MAXN], idx[MAXN];
long long dep[MAXN];
long long ad[MAXN * 4], mn[MAXN * 4];
struct edge {
  int u, v;
  long long w;
  int next;
} ed[MAXN * 2];
void dfs(int u) {
  st[u] = ++id;
  idx[id] = u;
  for (int i = head[u]; i != -1; i = ed[i].next) {
    dep[ed[i].v] = dep[u] + ed[i].w;
    dfs(ed[i].v);
  }
  edd[u] = id;
}
void build(int o, int l, int r) {
  if (l == r) {
    ad[o] = dep[idx[l]] + ed[path[idx[l]]].w;
    mn[o] = ad[o];
    return;
  }
  int mid = (l + r) / 2, lch = o * 2, rch = lch + 1;
  build(lch, l, mid);
  build(rch, mid + 1, r);
  mn[o] = min(mn[lch], mn[rch]);
}
void update(int o, int l, int r, int x, int y, long long v) {
  if (x <= l && r <= y) {
    ad[o] += v;
    mn[o] += v;
    return;
  }
  int mid = (l + r) / 2, lch = o * 2, rch = lch + 1;
  if (x <= mid) update(lch, l, mid, x, y, v);
  if (mid < y) update(rch, mid + 1, r, x, y, v);
  mn[o] = min(mn[lch], mn[rch]) + ad[o];
}
long long query(int o, int l, int r, int x, long long d) {
  if (l == r) return ad[o] + d;
  int mid = (l + r) / 2, lch = o * 2, rch = lch + 1;
  if (x <= mid)
    return query(lch, l, mid, x, d + ad[o]);
  else
    return query(rch, mid + 1, r, x, d + ad[o]);
}
long long querymn(int o, int l, int r, int x, int y, long long d) {
  if (x <= l && r <= y) return mn[o] + d;
  int mid = (l + r) / 2, lch = o * 2, rch = lch + 1;
  long long res = -1;
  if (x <= mid)
    res = res == -1 ? querymn(lch, l, mid, x, y, d + ad[o])
                    : min(res, querymn(lch, l, mid, x, y, d + ad[o]));
  if (mid < y)
    res = res == -1 ? querymn(rch, mid + 1, r, x, y, d + ad[o])
                    : min(res, querymn(rch, mid + 1, r, x, y, d + ad[o]));
  return res;
}
int main() {
  cin >> n >> q;
  memset(head, -1, sizeof(head));
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    long long w;
    scanf("%d%d%I64d", &u, &v, &w);
    ed[i] = {u, v, w, head[u]};
    head[u] = i;
  }
  for (int i = n; i <= 2 * n - 2; i++) {
    int u, v;
    long long w;
    scanf("%d%d%I64d", &u, &v, &w);
    ed[i] = {u, v, w, -1};
    path[u] = i;
  }
  dep[1] = 0;
  dfs(1);
  build(1, 1, n);
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int x;
      long long w;
      scanf("%d%I64d", &x, &w);
      if (x <= n - 1) {
        int v = ed[x].v;
        update(1, 1, n, st[v], edd[v], w - ed[x].w);
      } else {
        int v = ed[x].u;
        update(1, 1, n, st[v], st[v], w - ed[x].w);
      }
      ed[x].w = w;
    } else {
      int u, v;
      scanf("%d%d", &u, &v);
      if (u == v) {
        printf("0\n");
        continue;
      }
      long long du = query(1, 1, n, st[u], 0) - ed[path[u]].w,
                dv = query(1, 1, n, st[v], 0) - ed[path[v]].w;
      if (st[v] <= edd[u] && st[v] >= st[u])
        printf("%I64d\n", dv - du);
      else {
        long long t = querymn(1, 1, n, st[u], edd[u], 0);
        long long ans = t - du + dv;
        printf("%I64d\n", ans);
      }
    }
  }
}