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 N = 100005;
int n, m;
vector<int> G[N];
void addEdge(int u, int v) { G[u].push_back(v); }
struct Tree {
  int sumv[N << 2], addv[N << 2];
  void pushDown(int o, int L, int R) {
    if (addv[o] != 0) {
      int M = (L + R) / 2;
      sumv[(o << 1)] += addv[o] * (M - L + 1);
      sumv[(o << 1 | 1)] += addv[o] * (R - M);
      addv[(o << 1)] += addv[o];
      addv[(o << 1 | 1)] += addv[o];
      addv[o] = 0;
    }
  }
  void pushUp(int o) { sumv[o] = sumv[(o << 1)] + sumv[(o << 1 | 1)]; }
  int query(int o, int L, int R, int pos) {
    if (L == R) return sumv[o];
    int M = (L + R) / 2;
    pushDown(o, L, R);
    if (pos <= M)
      return query((o << 1), L, M, pos);
    else
      return query((o << 1 | 1), M + 1, R, pos);
  }
  void modify(int o, int L, int R, int ql, int qr, int val) {
    if (ql <= L && R <= qr) {
      sumv[o] += val * (R - L + 1);
      addv[o] += val;
      return;
    }
    int M = (L + R) / 2;
    pushDown(o, L, R);
    if (ql <= M) modify((o << 1), L, M, ql, qr, val);
    if (qr > M) modify((o << 1 | 1), M + 1, R, ql, qr, val);
    pushUp(o);
  }
} layer, chain;
int depth[N], son[N], id[N];
int dfs_clock;
void dfs(int u, int pre, int de) {
  id[u] = ++dfs_clock;
  depth[u] = de;
  son[u] = 0;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (pre == v) continue;
    dfs(v, u, de + 1);
    son[u] = son[v] + 1;
  }
}
int query(int v) {
  int ans = chain.query(1, 1, N, id[v]) + layer.query(1, 1, N, depth[v] + 1);
  return ans;
}
void modify(int v, int x, int d) {
  int length = d - depth[v];
  int l, r;
  if (length > 0) {
    layer.modify(1, 1, N, 1, length + 1, x);
    r = (d > son[v]) ? id[v] + son[v] : id[v] + d;
    if (v != 1) {
      l = id[v] + length - depth[v] + 1;
      chain.modify(1, 1, N, l, r, x);
    }
  } else {
    l = id[v] - d;
    r = (d > son[v]) ? id[v] + son[v] : id[v] + d;
    if (v != 1) {
      if (depth[v] - d == 0) {
        chain.modify(1, 1, N, l + 1, r, x);
        chain.modify(1, 1, N, 1, 1, x);
      } else {
        chain.modify(1, 1, N, l, r, x);
      }
    }
  }
}
int main() {
  int u, v;
  int oper, x, d;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &u, &v);
    addEdge(u, v);
    addEdge(v, u);
  }
  dfs(1, -1, 0);
  for (int i = 0; i < m; i++) {
    scanf("%d", &oper);
    if (oper == 1) {
      scanf("%d", &v);
      printf("%d\n", query(v));
    } else {
      scanf("%d%d%d", &v, &x, &d);
      modify(v, x, d);
    }
  }
  return 0;
}
