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;
const int mN = 1e5 + 100;
int n, q, ID[mN], mxb[mN], t, depth[mN], d;
vector<int> adjList[mN];
int dfs(int cur, int fr) {
  depth[cur] = d++;
  mxb[cur] = ID[cur] = t++;
  for (int i = 0; i < adjList[cur].size(); ++i) {
    if (adjList[cur][i] == fr) continue;
    mxb[cur] = max(mxb[cur], dfs(adjList[cur][i], cur));
  }
  d--;
  return mxb[cur];
}
int BIT[2][mN];
inline int get(int i, int z) {
  ++i;
  int ret = 0;
  while (i) ret += BIT[z][i], i -= i & -i;
  return ret;
}
inline void add(int i, int v, int z) {
  ++i;
  while (i < mN) BIT[z][i] += v, i += i & -i;
}
int main() {
  cin >> n >> q;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    scanf("%d", &u), scanf("%d", &v);
    adjList[u - 1].push_back(v - 1);
    adjList[v - 1].push_back(u - 1);
  }
  dfs(0, -1);
  for (int i = 0; i < q; ++i) {
    int t, v, x, d;
    scanf("%d", &t), scanf("%d", &v);
    --v;
    if (!t) {
      scanf("%d", &x), scanf("%d", &d);
      if (!v) {
        add(0, x, 0);
        add(d + 1, -x, 0);
        continue;
      }
      int l = max(ID[v] - d, ID[v] - depth[v] + 1), r = min(ID[v] + d, mxb[v]);
      if (d >= depth[v]) {
        add(0, x, 0);
        add((d - depth[v]) + 1, -x, 0);
        l = ID[v] - (2 * depth[v] - d - 1);
      }
      if (l > r) continue;
      add(l, x, 1);
      add(r + 1, -x, 1);
    } else {
      printf("%d\n", get(depth[v], 0) + get(ID[v], 1));
    }
  }
};
