A little girl loves problems on trees very much. Here's one of them.

A tree is an undirected connected graph, not containing cycles. The degree of node x in the tree is the number of nodes y of the tree, such that each of them is connected with node x by some edge of the tree. 

Let's consider a tree that consists of n nodes. We'll consider the tree's nodes indexed from 1 to n. The cosidered tree has the following property: each node except for node number 1 has the degree of at most 2.

Initially, each node of the tree contains number 0. Your task is to quickly process the requests of two types:

  * Request of form: 0 v x d. In reply to the request you should add x to all numbers that are written in the nodes that are located at the distance of at most d from node v. The distance between two nodes is the number of edges on the shortest path between them. 
  * Request of form: 1 v. In reply to the request you should print the current number that is written in node v. 

Input

The first line contains integers n (2 ≤ n ≤ 105) and q (1 ≤ q ≤ 105) — the number of tree nodes and the number of requests, correspondingly.

Each of the next n - 1 lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), that show that there is an edge between nodes ui and vi. Each edge's description occurs in the input exactly once. It is guaranteed that the given graph is a tree that has the property that is described in the statement.

Next q lines describe the requests.

  * The request to add has the following format: 0 v x d (1 ≤ v ≤ n, 1 ≤ x ≤ 104, 1 ≤ d < n). 
  * The request to print the node value has the following format: 1 v (1 ≤ v ≤ n). 



The numbers in the lines are separated by single spaces.

Output

For each request to print the node value print an integer — the reply to the request.

Examples

Input

3 6
1 2
1 3
0 3 1 2
0 2 3 1
0 1 5 2
1 1
1 2
1 3


Output

9
9
6


Input

6 11
1 2
2 5
5 4
1 6
1 3
0 3 1 3
0 3 4 5
0 2 1 4
0 1 5 5
0 4 6 2
1 1
1 2
1 3
1 4
1 5
1 6


Output

11
17
11
16
17
11
#include <bits/stdc++.h>
#pragma GCC optimize("O2")
long long poww(long long a, long long b, long long md) {
  return (!b ? 1
             : (b & 1 ? a * poww(a * a % md, b / 2, md) % md
                      : poww(a * a % md, b / 2, md) % md));
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
using namespace std;
const int maxn = 1000 * 100 + 5;
const long long inf = 9223372036854775807;
const long long mod = 1e9 + 7;
const long long Log = 22;
struct segtree {
  int sz = 1;
  vector<int> seg;
  void init(int n) {
    while (sz <= n) sz *= 2;
    seg.assign(2 * sz, 0);
  }
  void upd(int v, int val, int x, int lx, int rx) {
    if (rx - lx == 1) {
      seg[x] += val;
      return;
    }
    int mid = lx + rx >> 1;
    if (v < mid)
      upd(v, val, x << 1, lx, mid);
    else
      upd(v, val, x << 1 | 1, mid, rx);
    seg[x] = seg[x << 1] + seg[x << 1 | 1];
  }
  void upd(int v, int val) { return upd(v, val, 1, 0, sz); }
  int get(int l, int r, int x, int lx, int rx) {
    if (lx >= l && rx <= r) {
      return seg[x];
    }
    if (lx >= r || rx <= l) {
      return 0;
    }
    int mid = lx + rx >> 1;
    return get(l, r, x << 1, lx, mid) + get(l, r, x << 1 | 1, mid, rx);
  }
  int get(int l, int r) { return get(l, r, 1, 0, sz); }
};
int n, q, mark[maxn], sid[maxn], h[maxn], nh[maxn], cnt = 1, s[maxn], ans1;
vector<int> adj[maxn];
void dfs(int v) {
  mark[v] = 1;
  for (auto u : adj[v]) {
    if (!mark[u]) {
      if (v == 1) {
        sid[u] = cnt;
        cnt++;
      } else
        sid[u] = sid[v];
      h[u] = h[v] + 1;
      nh[h[u]]++;
      s[sid[u]] = max(s[sid[u]], h[u]);
      dfs(u);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> q;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1);
  segtree b[cnt], st;
  st.init(n);
  for (int i = 1; i < cnt; i++) {
    b[i].init(s[i]);
  }
  while (q--) {
    int t;
    cin >> t;
    if (t == 0) {
      int v, x, d;
      cin >> v >> x >> d;
      if (v == 1) {
        ans1 += x;
        st.upd(1, x);
        if (d <= n) st.upd(d - h[v] + 1, -1 * x);
        continue;
      }
      if (d - h[v] <= 0) {
        if (d - h[v] == 0) ans1 += x;
        int l = max(1, h[v] - d);
        b[sid[v]].upd(l, x);
        if (h[v] + d < s[sid[v]]) {
          b[sid[v]].upd(h[v] + d + 1, -1 * x);
        }
      } else {
        ans1 += x;
        int l = d - h[v] + 1;
        if (l <= s[sid[v]]) b[sid[v]].upd(l, x);
        if (h[v] + d < s[sid[v]]) {
          b[sid[v]].upd(h[v] + d + 1, -1 * x);
        }
        st.upd(1, x);
        if (d - h[v] + 1 <= n) {
          st.upd(d - h[v] + 1, -1 * x);
        }
      }
    } else {
      int v;
      cin >> v;
      if (v == 1)
        cout << ans1 << endl;
      else {
        cout << b[sid[v]].get(1, h[v] + 1) + st.get(1, h[v] + 1) << endl;
      }
    }
  }
}
