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 N = 800000 + 10;
const long long INF = 1e18;
int n, m, u[N], v[N], w[N];
int dfnL[N], dfnR[N], who[N], moment;
long long dis[N], lazy[N], val[N];
long long disToRoot[N];
vector<pair<int, int> > g[N];
long long par[N][20], dep[N];
void dfs(int u, int p) {
  dfnL[u] = ++moment;
  who[moment] = u;
  dep[u] = dep[p] + 1;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i].first;
    par[v][0] = u;
    dis[v] = dis[u] + g[u][i].second;
    dfs(v, u);
  }
  dfnR[u] = moment;
}
void push_down(int rt) {
  if (lazy[rt]) {
    lazy[rt << 1] += lazy[rt];
    lazy[rt << 1 | 1] += lazy[rt];
    val[rt << 1] += lazy[rt];
    val[rt << 1 | 1] += lazy[rt];
    lazy[rt] = 0;
  }
}
void build(int l, int r, int rt) {
  if (l == r) {
    val[rt] = dis[who[l]] + disToRoot[who[l]];
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, rt << 1);
  build(mid + 1, r, rt << 1 | 1);
  val[rt] = min(val[rt << 1], val[rt << 1 | 1]);
}
long long queryMin(int l, int r, int rt, int L, int R) {
  if (L <= l && r <= R) return val[rt];
  push_down(rt);
  int mid = (l + r) >> 1;
  long long ans = 1e16;
  if (L <= mid) ans = min(ans, queryMin(l, mid, rt << 1, L, R));
  if (R > mid) ans = min(ans, queryMin(mid + 1, r, rt << 1 | 1, L, R));
  return ans;
}
void update(int l, int r, int rt, int L, int R, long long x) {
  if (L <= l && r <= R) {
    lazy[rt] += x;
    val[rt] += x;
    return;
  }
  push_down(rt);
  int mid = (l + r) >> 1;
  if (L <= mid) update(l, mid, rt << 1, L, R, x);
  if (R > mid) update(mid + 1, r, rt << 1 | 1, L, R, x);
  val[rt] = min(val[rt << 1], val[rt << 1 | 1]);
}
bool ispar(int u, int v) {
  int k = dep[v] - dep[u];
  if (k < 0) return 0;
  int cur = v;
  long long ans = 0;
  for (int i = 0; i < 20; i++) {
    if ((k >> i) & 1) cur = par[cur][i];
  }
  return cur == u;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i < n; i++) {
    scanf("%d%d%d", &u[i], &v[i], &w[i]);
    g[u[i]].push_back(make_pair(v[i], w[i]));
  }
  for (int i = n; i < 2 * n - 1; i++) {
    scanf("%d%d%d", &u[i], &v[i], &w[i]);
    disToRoot[u[i]] = w[i];
  }
  dfs(1, 1);
  for (int i = 1; i < 20; i++) {
    for (int j = 1; j <= n; j++) {
      par[j][i] = par[par[j][i - 1]][i - 1];
    }
  }
  build(1, n, 1);
  for (int i = 1; i <= m; i++) {
    int op, x, y;
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1) {
      if (x < n) {
        update(1, n, 1, dfnL[v[x]], dfnR[v[x]], y - w[x]);
        w[x] = y;
      } else {
        update(1, n, 1, dfnL[u[x]], dfnL[u[x]], y - disToRoot[u[x]]);
        disToRoot[u[x]] = y;
      }
    }
    if (op == 2) {
      long long ans = queryMin(1, n, 1, dfnL[x], dfnR[x]) -
                      (queryMin(1, n, 1, dfnL[x], dfnL[x]) - disToRoot[x]) +
                      (queryMin(1, n, 1, dfnL[y], dfnL[y]) - disToRoot[y]);
      if (ispar(x, y)) {
        ans =
            min(ans, (queryMin(1, n, 1, dfnL[y], dfnL[y]) - disToRoot[y]) -
                         (queryMin(1, n, 1, dfnL[x], dfnL[x]) - disToRoot[x]));
      }
      printf("%lld\n", ans);
    }
  }
}