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>
int root[100010];
int node[100010][3];
int label[100010];
int depth[100010];
int sons[100010];
int base[100010];
int tree[100010];
int nodeone = 0;
inline int lowbit(int a) { return a & (-a); }
int update(int tree[], int a, int b, int n) {
  while (a <= n) {
    tree[a] += b;
    a += lowbit(a);
  }
  return 0;
}
int ask(int tree[], int a) {
  int resu = 0;
  while (a > 0) {
    resu += tree[a];
    a -= lowbit(a);
  }
  return resu;
}
std::stack<int> sta;
int dfs(int current) {
  sta.push(current);
  int flag;
  do {
    flag = 0;
    current = sta.top();
    for (int i = 1; i <= node[current][0]; i++)
      if (!label[node[current][i]]) {
        label[0]++;
        label[node[current][i]] = label[0];
        depth[node[current][i]] = depth[current] + 1;
        sta.push(node[current][i]);
        flag = 1;
      }
  } while (flag);
  int tag;
  while (!sta.empty()) {
    current = sta.top();
    if (node[current][0] == 1)
      tag = sons[current] = label[current];
    else
      sons[current] = tag;
    sta.pop();
  }
  return tag;
}
int main() {
  int n, m;
  while (scanf("%d%d", &n, &m) == 2) {
    memset(node, 0, sizeof(node));
    memset(root, 0, sizeof(root));
    for (int i = 1; i < n; i++) {
      int s, t;
      scanf("%d%d", &s, &t);
      if (s == 1)
        root[++root[0]] = t;
      else
        node[s][++node[s][0]] = t;
      if (t == 1)
        root[++root[0]] = s;
      else
        node[t][++node[t][0]] = s;
    }
    memset(label, 0, sizeof(label));
    memset(depth, 0, sizeof(depth));
    memset(sons, 0, sizeof(sons));
    memset(base, 0, sizeof(base));
    memset(tree, 0, sizeof(tree));
    label[0] = 1;
    label[1] = 1;
    depth[1] = 0;
    base[0] = 0;
    for (int i = 1; i <= root[0]; i++) {
      label[root[i]] = ++label[0];
      depth[root[i]] = 1;
      dfs(root[i]);
      if (base[0] < sons[root[i]] - label[root[i]] + 1)
        base[0] = sons[root[i]] - label[root[i]] + 1;
    }
    nodeone = 0;
    for (int i = 1; i <= m; i++) {
      int flag;
      scanf("%d", &flag);
      if (flag == 1) {
        int t;
        scanf("%d", &t);
        if (t == 1)
          printf("%d\n", nodeone);
        else {
          int temp = ask(tree, label[t]) + ask(base, depth[t]);
          printf("%d\n", temp);
        }
      } else if (flag == 0) {
        int v, x, d;
        scanf("%d%d%d", &v, &x, &d);
        if (v == 1) {
          nodeone += x;
          int s = 1, t = d;
          update(base, s, x, base[0]);
          update(base, t + 1, -x, base[0]);
        } else {
          if (d < depth[v]) {
            int s = label[v] - d;
            int t = label[v] + d;
            if (t > sons[v]) t = sons[v];
            update(tree, s, x, n);
            update(tree, t + 1, -x, n);
          } else if (d == depth[v]) {
            nodeone += x;
            int s = label[v] - d + 1;
            int t = label[v] + d;
            if (t > sons[v]) t = sons[v];
            update(tree, s, x, n);
            update(tree, t + 1, -x, n);
          } else {
            nodeone += x;
            int s1 = 1, t1 = d - depth[v];
            if (t1 > base[0]) t1 = base[0];
            update(base, s1, x, base[0]);
            update(base, t1 + 1, -x, base[0]);
            int orig = label[v] - depth[v] + 1;
            int s2 = orig + t1;
            if (s2 > sons[v]) continue;
            int t2 = label[v] + d;
            if (t2 > sons[v]) t2 = sons[v];
            update(tree, s2, x, n);
            update(tree, t2 + 1, -x, n);
          }
        }
      }
    }
  }
  fclose(stdin);
  fclose(stdout);
  return 0;
}
