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>
const int INF = 0x3f3f3f3f;
void qread(int &x) {
  x = 0;
  char c = getchar();
  while (c < '0' or c > '9') c = getchar();
  while (c >= '0' and c <= '9') x = x * 10 + c - '0', c = getchar();
}
void qreadll(long long &x) {
  x = 0;
  char c = getchar();
  while (c < '0' or c > '9') c = getchar();
  while (c >= '0' and c <= '9') x = x * 10ll + c - '0', c = getchar();
}
long long mymin(long long x, long long y) { return (x < y) ? x : y; }
const int MAXN = 210000;
struct Seg {
  int l, r, mid;
  int lc, rc;
  long long mi, lz;
} s[MAXN * 2];
int cnt = 0;
int build(int l, int r) {
  int t = ++cnt;
  if (l == r)
    s[t] = (Seg){l, r, l, 0, 0, 0, 0};
  else {
    s[t] = (Seg){l, r, (l + r) >> 1, 0, 0, 0, 0};
    s[t].lc = build(l, s[t].mid);
    s[t].rc = build(s[t].mid + 1, r);
    s[t].mi = mymin(s[s[t].lc].mi, s[s[t].rc].mi);
  }
  return t;
}
void pushdown(int x) {
  int lc = s[x].lc, rc = s[x].rc;
  if (lc > 0) {
    s[lc].mi += s[x].lz;
    s[lc].lz += s[x].lz;
  }
  if (rc > 0) {
    s[rc].mi += s[x].lz;
    s[rc].lz += s[x].lz;
  }
  s[x].lz = 0;
}
void change(int x, int l, int r, long long o) {
  if (s[x].l == l and s[x].r == r) {
    s[x].mi += o;
    s[x].lz += o;
    return;
  }
  if (s[x].lz != 0) pushdown(x);
  int lc = s[x].lc, rc = s[x].rc;
  if (r <= s[x].mid)
    change(lc, l, r, o);
  else if (l > s[x].mid)
    change(rc, l, r, o);
  else
    change(lc, l, s[x].mid, o), change(rc, s[x].mid + 1, r, o);
  s[x].mi = mymin(s[lc].mi, s[rc].mi);
}
long long ask(int x, int l, int r) {
  if (s[x].l == l and s[x].r == r) return s[x].mi;
  if (s[x].lz != 0) pushdown(x);
  int lc = s[x].lc, rc = s[x].rc;
  if (r <= s[x].mid) return ask(lc, l, r);
  if (l > s[x].mid) return ask(rc, l, r);
  return mymin(ask(lc, l, s[x].mid), ask(rc, s[x].mid + 1, r));
}
struct Nod {
  int hou;
  int siz;
  long long dis, tt;
  Nod() { hou = 0; }
} p[MAXN];
struct Edge {
  int y, g;
  long long c;
} e[MAXN * 2];
int ln = 0;
void ins(int x, int y, long long c) {
  e[++ln] = (Edge){y, p[x].hou, c};
  p[x].hou = ln;
}
int yz[MAXN], id = 0;
void dfs(int x, int fa) {
  yz[x] = ++id;
  p[x].siz = 1;
  for (int k = p[x].hou; k > 0; k = e[k].g) {
    int y = e[k].y;
    if (y == fa) continue;
    p[y].dis = p[x].dis + e[k].c;
    dfs(y, x);
    p[x].siz += p[y].siz;
  }
}
struct SG {
  int x, y;
  long long z;
} ss[2 * MAXN];
long long solve(int x, int y) {
  if (yz[x] <= yz[y] and yz[y] <= yz[x] + p[x].siz - 1)
    return (ask(1, yz[y], yz[y]) - p[y].tt) - (ask(1, yz[x], yz[x]) - p[x].tt);
  return ask(1, yz[x], yz[x] + p[x].siz - 1) -
         (ask(1, yz[x], yz[x]) - p[x].tt) + (ask(1, yz[y], yz[y]) - p[y].tt);
}
int main() {
  int n, q;
  qread(n);
  qread(q);
  for (int i = 1; i <= 2 * n - 2; i++) {
    int x, y;
    long long z;
    qread(x);
    qread(y);
    qreadll(z);
    ss[i] = (SG){x, y, z};
    if (y == 1)
      p[x].tt = z;
    else
      ins(x, y, z);
  }
  dfs(1, 0);
  build(1, n);
  for (int i = 2; i <= n; i++) change(1, yz[i], yz[i], p[i].dis + p[i].tt);
  while (q--) {
    int id, x, y;
    qread(id);
    qread(x);
    qread(y);
    if (id == 2)
      printf("%I64d\n", solve(x, y));
    else {
      int nx = ss[x].x, ny = ss[x].y;
      if (ny == 1)
        change(1, yz[nx], yz[nx], -p[nx].tt + y), p[nx].tt = y;
      else
        change(1, yz[ny], yz[ny] + p[ny].siz - 1, -ss[x].z + y), ss[x].z = y;
    }
  }
}