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;
template <typename T>
inline void Int(T &n) {
  n = 0;
  int f = 1;
  register int ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) n = (n << 3) + (n << 1) + ch - '0';
  n = n * f;
}
template <typename T>
T gcd(T a, T b) {
  return !b ? a : gcd(b, a % b);
}
template <typename T>
inline void umin(T &a, T b) {
  a = a < b ? a : b;
}
template <typename T>
inline void umax(T &a, T b) {
  a = a > b ? a : b;
}
template <typename T, typename W>
inline void Int(T &x, W &y) {
  Int(x), Int(y);
}
template <typename T, typename W, typename Q>
inline void Int(T &x, W &y, Q &z) {
  Int(x, y), Int(z);
}
const int N = 1e5 + 7;
const int inf = 1e9 + 7;
vector<int> g[N], fen[N];
int chainId, lev[N], sz[N], belongs[N];
void modify(int x, int y, int z) {
  while (y <= sz[x]) fen[x][y] += z, y += y & -y;
}
int ask(int x, int y) {
  int res = 0;
  while (y > 0) res += fen[x][y], y -= y & -y;
  return res;
}
void dfs(int u, int p) {
  lev[u] = lev[p] + 1;
  sz[chainId]++;
  belongs[u] = chainId;
  for (int i = 1; i <= g[u].size(); ++i) {
    int v = g[u][i - 1];
    if (v != p) dfs(v, u);
  }
}
int solve() {
  int n, m;
  Int(n, m);
  for (int i = 2; i <= n; ++i) {
    int u, v;
    Int(u, v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i <= n + 2; ++i) sz[i] = 3;
  lev[1] = 1, belongs[1] = n + 1;
  for (int v : g[1]) {
    ++chainId;
    dfs(v, 1);
  }
  sz[0] = n + 2;
  fen[0].resize(n + 5);
  for (int i = 1; i <= n + 2; ++i) fen[i].resize(sz[i] + 3);
  while (m--) {
    int t, v, x, d;
    Int(t, v);
    int vchain = belongs[v], R, L;
    if (t == 0) {
      Int(x, d);
      if (v == 1) {
        modify(0, 1, x);
        modify(0, min(sz[0], d + lev[v] + 1), -x);
        continue;
      }
      L = max(2, lev[v] - d);
      R = min(sz[vchain], lev[v] + d);
      modify(vchain, L, x);
      modify(vchain, R + 1, -x);
      if (lev[v] - d == 1) {
        modify(0, 1, x);
        modify(0, 2, -x);
      } else if (lev[v] - d < 1) {
        modify(0, 1, x);
        modify(0, min(sz[0], d - lev[v] + 3), -x);
        modify(vchain, 2, -x);
        modify(vchain, min(sz[vchain], d - lev[v] + 3), x);
      }
    } else {
      printf("%d\n", ask(0, lev[v]) + ask(vchain, lev[v]));
    }
  }
  return 0;
}
int main() {
  int tests = 1, CaseNo = 0;
  while (tests--) {
    solve();
  }
  return 0;
}
