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;
vector<int> bit[100050];
int root[100050];
int depth[100050];
int chain[100050];
int par[100050];
vector<int> g[100050];
int n;
void dfs() {
  int x = 1;
  depth[1] = 0;
  chain[1] = 0;
  for (int i = 0; i < g[1].size(); i++) {
    int p = g[1][i];
    int d = 1;
    par[p] = 1;
    depth[p] = 1;
    chain[p] = x;
    while (g[p].size() == 2) {
      int y = g[p][0];
      if (y == par[p]) y = g[p][1];
      d++;
      depth[y] = d;
      chain[y] = x;
      par[y] = p;
      p = y;
    }
    bit[x].resize(d + 50);
    x++;
  }
}
void update(int x, int idx, int val) {
  while (idx < bit[x].size()) {
    bit[x][idx] += val;
    idx += (idx & (-idx));
  }
  return;
}
int read(int x, int idx) {
  int sum = 0;
  while (idx > 0) {
    sum += bit[x][idx];
    idx -= (idx & (-idx));
  }
  return sum;
}
void up_root(int d, int val) {
  while (d <= 100005) {
    root[d] += val;
    d += (d & -d);
  }
}
int rd_root(int d) {
  int sum = 0;
  while (d > 0) {
    sum += root[d];
    d -= (d & -d);
  }
  return sum;
}
int main() {
  int q;
  cin >> n >> q;
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs();
  for (int i = 0; i < q; i++) {
    int t;
    scanf("%d", &t);
    if (t == 0) {
      int x, y, z;
      scanf("%d %d %d", &x, &z, &y);
      if (x == 1) {
        up_root(y + 1 + 1, -z);
        up_root(1, z);
      } else {
        update(chain[x], depth[x], z);
        update(chain[x], depth[x] + y + 1, -z);
        int d = y;
        int r = x;
        d = y - depth[x];
        if (d < 0) {
          update(chain[r], depth[x] - y, z);
          update(chain[r], depth[r], -z);
        } else {
          up_root(1, z);
          up_root(d + 1 + 1, -z);
          update(chain[r], d + 1, z);
          update(chain[r], depth[r], -z);
        }
      }
    } else {
      int x;
      scanf("%d", &x);
      int res = rd_root(depth[x] + 1);
      if (x != 1) res += read(chain[x], depth[x]);
      printf("%d\n", res);
    }
  }
  return 0;
}
