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 maxn = 1e5 + 100;
vector<int> a[maxn];
int h[maxn];
int sub[maxn];
int en[maxn];
int val[4 * maxn];
vector<int> valc[maxn];
void dfs(int x, int par, int su) {
  sub[x] = su;
  for (int i = 0; i < a[x].size(); i++) {
    int u = a[x][i];
    if (u == par) continue;
    h[u] = h[x] + 1;
    if (x == 1) {
      dfs(u, x, i + 1);
      continue;
    }
    dfs(u, x, su);
  }
  if (a[x].size() == 1 and x != 1) en[su] = h[x];
  return;
}
void add(int id, int L, int R, int l, int r, int ty, int x) {
  if (l == L and r == R) {
    valc[ty][id] += x;
    return;
  }
  int mid = (L + R) >> 1;
  if (l < mid) add(id * 2 + 0, L, mid, l, min(r, mid), ty, x);
  if (r > mid) add(id * 2 + 1, mid, R, max(l, mid), r, ty, x);
  return;
}
void add2(int id, int L, int R, int l, int r, int x) {
  if (l == L and r == R) {
    val[id] += x;
    return;
  }
  int mid = (L + R) >> 1;
  if (l < mid) add2(id * 2 + 0, L, mid, l, min(r, mid), x);
  if (r > mid) add2(id * 2 + 1, mid, R, max(l, mid), r, x);
  return;
}
int get(int id, int L, int R, int idx, int ty) {
  int ret = valc[ty][id];
  if (L + 1 == R) return ret;
  int mid = (L + R) >> 1;
  if (idx < mid)
    ret += get(id * 2 + 0, L, mid, idx, ty);
  else
    ret += get(id * 2 + 1, mid, R, idx, ty);
  return ret;
}
int get1(int id, int L, int R, int idx) {
  int ret = val[id];
  if (L + 1 == R) return ret;
  int mid = (L + R) >> 1;
  if (idx < mid)
    ret += get1(id * 2 + 0, L, mid, idx);
  else
    ret += get1(id * 2 + 1, mid, R, idx);
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, q;
  cin >> n >> q;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    a[u].push_back(v);
    a[v].push_back(u);
  }
  dfs(1, 0, 0);
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= 10 * en[i]; j++) valc[i].push_back(0);
  for (int i = 1; i <= q; i++) {
    int ty, u, d, ki;
    cin >> ty >> u;
    if (ty == 0) {
      cin >> ki >> d;
      int su = sub[u];
      int tem = d - h[u];
      if (d >= h[u]) {
        if (u != 1 and tem < en[su])
          add(1, 1, en[su] + 1, tem + 1, min(en[su], h[u] + d) + 1, su, ki);
        add2(1, 1, n + 1, 1, tem + 2, ki);
      } else {
        add(1, 1, en[su] + 1, h[u] - d, min(en[su], h[u] + d) + 1, su, ki);
        continue;
      }
    } else if (ty == 1) {
      int su = sub[u];
      int ans = 0;
      if (u != 1) ans += get(1, 1, en[su] + 1, h[u], su);
      ans += get1(1, 1, n + 1, h[u] + 1);
      cout << ans << endl;
    }
  }
  return 0;
}
