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;
int n, m;
vector<int> e[100005];
int d[100005], f[100005];
struct Fenwick {
  vector<int> a;
  int n;
  void add(int x, int y) {
    for (int i = x; i <= n; i += i & -i) {
      a[i] += y;
    }
  }
  int get(int x) {
    int s = 0;
    for (int i = x; i > 0; i -= i & -i) {
      s += a[i];
    }
    return s;
  }
} cc;
Fenwick c[100005];
void dfs(int k, int fa) {
  if (fa != -1) {
    f[k] = f[fa];
  }
  if (fa == 0) {
    f[k] = k;
  }
  for (int i = 0; i < e[k].size(); ++i) {
    int u = e[k][i];
    if (u != fa) {
      d[u] = d[k] + 1;
      dfs(u, k);
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  int x, y, z;
  for (int i = 0; i < n - 1; ++i) {
    scanf("%d%d", &x, &y);
    --x;
    --y;
    e[x].push_back(y);
    e[y].push_back(x);
  }
  d[0] = 0;
  dfs(0, -1);
  for (int i = 1; i < n; ++i) {
    c[f[i]].a.push_back(0);
    if (f[i] == i) {
      c[f[i]].a.push_back(0);
    }
  }
  for (int i = 1; i < n; ++i) {
    c[f[i]].n = c[f[i]].a.size() - 1;
  }
  for (int i = 1; i <= n + 1; ++i) {
    cc.a.push_back(0);
  }
  cc.n = cc.a.size() - 1;
  int ans = 0;
  for (int i = 0; i < m; ++i) {
    scanf("%d", &x);
    if (x == 0) {
      scanf("%d%d%d", &x, &y, &z);
      --x;
      if (x == 0) {
        ans += y;
        cc.add(1, y);
        cc.add(z + 1, -y);
        continue;
      }
      if (d[x] >= z) {
        c[f[x]].add(max(d[x] - z, 1), y);
        if (d[x] + z + 1 <= c[f[x]].n) {
          c[f[x]].add(d[x] + z + 1, -y);
        }
        if (d[x] == z) {
          ans += y;
        }
      } else {
        ans += y;
        if (z - d[x] + 1 <= c[f[x]].n) {
          c[f[x]].add(z - d[x] + 1, y);
        }
        if (d[x] + z + 1 <= c[f[x]].n) {
          c[f[x]].add(d[x] + z + 1, -y);
        }
        cc.add(1, y);
        cc.add(z - d[x] + 1, -y);
      }
    } else {
      scanf("%d", &x);
      --x;
      if (x == 0) {
        printf("%d\n", ans);
      } else {
        printf("%d\n", cc.get(d[x]) + c[f[x]].get(d[x]));
      }
    }
  }
  return 0;
}
