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;
char ch_ch_ch[1 << 20];
inline int gi() {
  int a;
  scanf("%d", &a);
  return a;
}
inline string gs() {
  scanf("%s", ch_ch_ch);
  return string(ch_ch_ch);
}
inline string gl() {
  gets(ch_ch_ch);
  return string(ch_ch_ch);
}
const int inf = 2000000000;
int inTree[100100];
int depth[100100];
vector<vector<int> > g;
vector<vector<int> > t;
void dfs(int v, int d, int num, int p) {
  inTree[v] = num;
  depth[v] = d;
  t[num].push_back(0);
  for (int i = (0); i < (g[v].size()); ++i) {
    int to = g[v][i];
    if (to == p) continue;
    dfs(g[v][i], d + 1, num, v);
  }
}
void update(int p, int v, int num) {
  for (int i = p; i < t[num].size(); i += (i & (-i))) t[num][i] += v;
}
int find(int p, int num) {
  int res = 0;
  for (int i = p; i > 0; i -= (i & (-i))) res += t[num][i];
  return res;
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  g.resize(n);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  t.resize(g[0].size() + 1);
  int inRoot = 0;
  for (int i = (0); i < (g[0].size()); ++i) {
    t[i].push_back(0);
    dfs(g[0][i], 1, i, 0);
  }
  t[t.size() - 1].resize(n + 10);
  for (int i = (0); i < (q); ++i) {
    int ty;
    scanf("%d", &ty);
    if (ty == 0) {
      int v, val, dist;
      scanf("%d%d%d", &v, &val, &dist);
      v--;
      if (v == 0) {
        inRoot += val;
        update(1, val, t.size() - 1);
        update(dist + 1, -val, t.size() - 1);
      } else {
        if (dist >= depth[v]) {
          int left = dist - depth[v];
          inRoot += val;
          update(1, val, t.size() - 1);
          update(left + 1, -val, t.size() - 1);
          update(left + 1, val, inTree[v]);
          update(depth[v] + dist + 1, -val, inTree[v]);
        } else {
          update(depth[v] - dist, val, inTree[v]);
          update(depth[v] + dist + 1, -val, inTree[v]);
        }
      }
    } else {
      int v;
      scanf("%d", &v);
      v--;
      if (v == 0)
        printf("%d\n", inRoot);
      else {
        int res = find(depth[v], inTree[v]) + find(depth[v], t.size() - 1);
        printf("%d\n", res);
      }
    }
  }
  return 0;
}
