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>
using namespace std;
template <class T>
void RD(T &a) {
  register int ch = getchar();
  bool ng = 0;
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') ng = 1;
  for (a = 0; ch >= '0' && ch <= '9'; ch = getchar())
    a = (a << 1) + (a << 3) + (ch ^ '0');
  if (ng) a = ~a + 1;
}
template <class T>
inline void WT(T a) {
  if (a < 0)
    putchar('-'), WT(~a + 1);
  else {
    if (a > 9) WT(a / 10);
    putchar(a % 10 ^ '0');
  }
}
vector<vector<int> > T;
inline void modify(int id, int p, int x) {
  for (; p < (int)((T[id]).size()); p += p & (-p)) {
    T[id][p] += x;
  }
}
inline void upd(int id, int l, int r, int x) {
  modify(id, l, x), modify(id, r + 1, -x);
}
inline int get(int id, int p) {
  int res = 0;
  for (; p; p -= p & (-p)) {
    res += T[id][p];
  }
  return res;
}
const int MAXN = 1e5 + 5;
int n, q, ind[MAXN], sz[MAXN], h[MAXN];
vector<int> G[MAXN];
inline void dfs(int u, int p) {
  for (int i(0), b_((int)((G[u]).size()) - 1); i <= b_; ++i) {
    int v = G[u][i];
    if (v == p) continue;
    ind[v] = ind[u], ++sz[ind[v]];
    h[v] = h[u] + 1, dfs(v, u);
  }
}
int main() {
  RD(n), RD(q);
  for (int i(1), b_(n - 1); i <= b_; ++i) {
    int u, v;
    RD(u), RD(v);
    G[u].emplace_back(v);
    G[v].emplace_back(u);
  }
  T.resize((int)((G[1]).size()) + 2);
  for (int i(0), b_((int)((G[1]).size()) - 1); i <= b_; ++i) {
    ind[G[1][i]] = i + 1;
    sz[i + 1] = h[G[1][i]] = 1;
    dfs(G[1][i], 1);
  }
  for (int i(1), b_((int)((G[1]).size())); i <= b_; ++i) {
    T[i].assign(sz[i] + 1, 0);
  }
  int mxh = 0;
  for (int i(1), b_(n); i <= b_; ++i) mxh = max(mxh, h[i]);
  T[(int)((G[1]).size()) + 1].assign(mxh + 1, 0);
  int root = 0;
  while (q--) {
    int t;
    RD(t);
    if (!t) {
      int v, d, x;
      RD(v), RD(x), RD(d);
      if (h[v] > d) {
        upd(ind[v], h[v] - d, min(h[v] + d, sz[ind[v]]), x);
      } else if (h[v] == d) {
        root += x;
        upd(ind[v], 1, min(h[v] + d, sz[ind[v]]), x);
      } else {
        root += x;
        upd((int)((G[1]).size()) + 1, 1, min(d - h[v], mxh), x);
        upd(ind[v], d - h[v] + 1, min(h[v] + d, sz[ind[v]]), x);
      }
    } else {
      int v;
      RD(v);
      if (v == 1)
        WT(root);
      else
        WT(get((int)((G[1]).size()) + 1, h[v]) + get(ind[v], h[v]));
      puts("");
    }
  }
  return 0;
}
