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>
const int MAXN = 222222;
const int MAXF = 222222;
int ss[MAXN], ff[MAXN], md[MAXN], vv[MAXN], id[MAXN], de[MAXN];
int fcaca[MAXN * 10];
int* f[MAXN];
int* allf;
int getSum(int* f, int m, int x) {
  int ret = 0;
  for (int i = x; i >= 0; i = (i & (i + 1)) - 1) {
    ret += f[i];
  }
  return ret;
}
void add2(int* f, int m, int x, int y) {
  for (int i = x; i < m; i |= i + 1) {
    f[i] += y;
  }
}
void addValue(int* f, int m, int x, int y) {
  for (int i = x; i >= 0; i = (i & (i + 1)) - 1) {
    f[i] += y;
  }
}
int getValue(int* f, int m, int x) {
  int ret = 0;
  for (int i = x; i < m; i |= i + 1) {
    ret += f[i];
  }
  return ret;
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 0; i <= n; i++) ss[i] = ff[i] = -1;
  int cf = 0;
  for (int i = 0; i + 1 < n; i++) {
    int v, u;
    scanf("%d %d", &v, &u);
    if (v != 1) {
      if (ss[v] >= 0)
        ff[v] = u;
      else
        ss[v] = u;
    } else {
      vv[cf++] = u;
    }
    if (u != 1) {
      if (ss[u] >= 0)
        ff[u] = v;
      else
        ss[u] = v;
    } else {
      vv[cf++] = v;
    }
  }
  int* fca = fcaca;
  allf = fca;
  fca = fca + (2 * MAXF);
  for (int it = 0; it < cf; it++) {
    int i = vv[it];
    int last = 1;
    int cnt = 1;
    while (i >= 0) {
      id[i] = it;
      de[i] = cnt++;
      int j = ss[i] ^ ff[i] ^ last;
      last = i;
      i = j;
    }
    md[it] = cnt + 1;
    f[it] = fca;
    fca = fca + cnt + 1;
  }
  for (int i = 0; i < q; i++) {
    int type, v;
    scanf("%d %d", &type, &v);
    if (type == 0) {
      int add, d;
      scanf("%d %d", &add, &d);
      if (v == 1) {
        add2(allf, MAXF, MAXF - d - 1, add);
      } else if (de[v] > d) {
        addValue(f[id[v]], md[id[v]], de[v] - d - 1, -add);
        int till = de[v] + d;
        if (till > md[id[v]] - 1) till = md[id[v]] - 1;
        addValue(f[id[v]], md[id[v]], till, add);
      } else {
        addValue(f[id[v]], md[id[v]], 0, -add);
        int till = de[v] + d;
        if (till > md[id[v]] - 1) till = md[id[v]] - 1;
        addValue(f[id[v]], md[id[v]], till, add);
        add2(allf, MAXF, MAXF - (d - de[v]) - 1, add);
        till = d - de[v];
        if (till > md[id[v]] - 1) till = md[id[v]] - 1;
        addValue(f[id[v]], md[id[v]], till, -add);
      }
    } else {
      int ans = v == 1 ? 0 : getValue(f[id[v]], md[id[v]], de[v]);
      printf("%d\n", ans + getSum(allf, MAXF, MAXF - de[v] - 1));
    }
  }
}
