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;
using ll = long long;
const ll N = 2e5 + 1, LOG = 20;
struct EDGE {
  ll u, v, val;
} edge[N << 1];
ll l, r, id, val;
ll tree[N << 2], lazy[N << 2], cost[N];
ll n, t, st[N], en[N];
ll arr[N], d[N], p[N][LOG];
vector<pair<ll, ll> > g[N];
void dfs(ll u = 1, ll par = 0) {
  st[u] = ++t;
  d[u] = d[par] + 1;
  p[u][0] = par;
  for (ll i = 1; i < LOG; i += 1) p[u][i] = p[p[u][i - 1]][i - 1];
  for (auto &z : g[u]) {
    ll v = z.first;
    arr[v] = arr[u] + z.second;
    dfs(v, u);
  }
  en[u] = t;
}
void propagate(ll node, ll b, ll e) {
  if (lazy[node]) {
    tree[node] += lazy[node];
    if (b != e) {
      lazy[node << 1] += lazy[node];
      lazy[node << 1 | 1] += lazy[node];
    }
    lazy[node] = 0;
  }
}
void update(ll node = 1, ll b = 1, ll e = n) {
  propagate(node, b, e);
  if (b > e || b > r || e < l) return;
  if (l <= b && e <= r) {
    lazy[node] += val;
    propagate(node, b, e);
    return;
  }
  ll m = (b + e) >> 1;
  update(node << 1, b, m);
  update(node << 1 | 1, m + 1, e);
  tree[node] = min(tree[node << 1], tree[node << 1 | 1]);
}
ll query(ll node = 1, ll b = 1, ll e = n) {
  propagate(node, b, e);
  if (b > e || b > r || e < l) return 1e18;
  if (l <= b && e <= r) return tree[node];
  ll m = (b + e) >> 1;
  return min(query(node << 1, b, m), query(node << 1 | 1, m + 1, e));
}
bool isAncestor(ll u, ll v) {
  if (d[u] > d[v]) return 0;
  swap(u, v);
  for (ll i = LOG - 1; ~i; i -= 1)
    if (d[p[u][i]] >= d[v]) u = p[u][i];
  return v == u;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ll q;
  cin >> n >> q;
  for (ll i = 1; i < n; i += 1) {
    cin >> edge[i].u >> edge[i].v >> edge[i].val;
    g[edge[i].u].push_back({edge[i].v, edge[i].val});
  }
  dfs();
  for (ll i = n; i < n + n - 1; i += 1) {
    cin >> edge[i].u >> edge[i].v >> edge[i].val;
    cost[edge[i].u] = edge[i].val;
  }
  for (int i = 1; i <= n; i += 1) {
    l = st[i], r = st[i], val = arr[i] + cost[i];
    update();
  }
  while (q--) {
    ll op;
    cin >> op;
    if (op == 1) {
      ll edgeno, x;
      cin >> edgeno >> x;
      ll u;
      if (edgeno < n) {
        u = edge[edgeno].v;
        l = st[u], r = en[u];
        val = x - edge[edgeno].val;
        edge[edgeno].val = x;
        update();
      } else {
        u = edge[edgeno].u;
        l = st[u], r = st[u];
        val = x - edge[edgeno].val;
        cost[u] = x;
        edge[edgeno].val = x;
        update();
      }
    } else {
      ll u, v;
      ll ans;
      cin >> u >> v;
      if (isAncestor(u, v)) {
        l = st[v], r = st[v], ans = query() - cost[v];
        l = st[u], r = st[u], ans -= (query() - cost[u]);
      } else {
        ll a, b, c;
        l = st[u], r = en[u], a = query();
        l = st[v], r = st[v], b = query() - cost[v];
        l = st[u], r = st[u], c = query() - cost[u];
        ans = a + b - c;
      }
      cout << ans << '\n';
    }
  }
  return 0;
}