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;
long long getint() {
  long long ans = 0;
  char c = getchar();
  while (c > '9' || c < '0') c = getchar();
  do {
    ans = ans * 10 + c - '0';
    c = getchar();
  } while (c >= '0' && c <= '9');
  return ans;
}
const long long N = 200010, M = N * 2;
long long n, q;
struct bian {
  long long l, e, n;
};
bian b[M];
long long s[N], tot = 0;
void add(long long x, long long y, long long z) {
  tot++;
  b[tot].l = z;
  b[tot].e = y;
  b[tot].n = s[x];
  s[x] = tot;
}
long long a[N << 2], tag[N << 2];
inline void pushup(long long x) { a[x] = min(a[x << 1], a[x << 1 | 1]); }
inline void pushdown(long long x) {
  tag[x << 1] += tag[x];
  tag[x << 1 | 1] += tag[x];
  a[x << 1] += tag[x];
  a[x << 1 | 1] += tag[x];
  tag[x] = 0;
}
inline void change(long long l, long long r, long long val, long long x,
                   long long nl, long long nr) {
  if (nl >= l && nr <= r) {
    tag[x] += val;
    a[x] += val;
    return;
  }
  pushdown(x);
  long long mid = nl + nr >> 1;
  if (nl <= r && mid >= l) change(l, r, val, x << 1, nl, mid);
  if (mid + 1 <= r && nr >= l) change(l, r, val, x << 1 | 1, mid + 1, nr);
  pushup(x);
}
inline long long ask(long long l, long long r, long long x, long long nl,
                     long long nr) {
  if (nl >= l && nr <= r) return a[x];
  pushdown(x);
  long long mid = nl + nr >> 1;
  long long ans = 0x7f7f7f7f;
  if (nl <= r && mid >= l) ans = min(ans, ask(l, r, x << 1, nl, mid));
  if (mid + 1 <= r && nr >= l)
    ans = min(ans, ask(l, r, x << 1 | 1, mid + 1, nr));
  return ans;
}
signed dfn[N], dfnend[N], dfnn = 1;
long long dis[N], dep[N];
signed fa[N], f[30][N];
bool check(long long u, long long v) {
  if (u == v) return 1;
  if (dep[u] >= dep[v]) return 0;
  for (long long i = 29; i >= 0; --i) {
    if (dep[f[i][v]] >= dep[u]) v = f[i][v];
  }
  return u == v;
}
long long ss(long long x) {
  dfnend[x] = dfn[x] = dfnn;
  dfnn++;
  for (long long i = s[x]; i; i = b[i].n) {
    dis[b[i].e] = dis[x] + b[i].l;
    f[0][b[i].e] = fa[b[i].e] = x;
    dep[b[i].e] = dep[x] + 1;
    dfnend[x] = ss(b[i].e);
  }
  return dfnend[x];
}
void initf() {
  for (long long i = 1; i < 30; i++) {
    for (long long j = 1; j <= n; j++) {
      f[i][j] = f[i - 1][f[i - 1][j]];
    }
  }
}
long long tofa[N], to1[N];
signed main() {
  n = getint(), q = getint();
  for (long long i = 0; i < n - 1; i++) {
    long long x = getint(), y = getint(), z = getint();
    add(x, y, z);
    tofa[y] = z;
  }
  dep[1] = 1;
  ss(1);
  initf();
  for (long long i = 0; i < n - 1; i++) {
    long long x = getint(), y = getint(), z = getint();
    add(y, x, z);
    to1[x] = z;
  }
  for (long long i = 1; i <= n; i++) {
    change(dfn[i], dfn[i], dis[i] + to1[i], 1, 1, n);
  }
  while (q--) {
    long long x = getint();
    if (x == 1) {
      long long x = getint(), y = getint();
      if (x <= n - 1) {
        change(dfn[b[x].e], dfnend[b[x].e], y - b[x].l, 1, 1, n);
        b[x].l = y;
      } else {
        change(dfn[b[x].e], dfn[b[x].e], y - b[x].l, 1, 1, n);
        b[x].l = to1[b[x].e] = y;
      }
    } else {
      long long u = getint(), v = getint();
      if (check(u, v)) {
        printf("%lld\n", (ask(dfn[v], dfn[v], 1, 1, n) - to1[v]) -
                             (ask(dfn[u], dfn[u], 1, 1, n) - to1[u]));
      } else {
        printf("%lld\n", ask(dfn[u], dfnend[u], 1, 1, n) +
                             (ask(dfn[v], dfn[v], 1, 1, n) - to1[v]) -
                             (ask(dfn[u], dfn[u], 1, 1, n) - to1[u]));
      }
    }
  }
}