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 MAX_N = 1000 * 100 + 10;
int n, qq, ans = 0;
long long tab = 0;
long long seg[4 * MAX_N], dist[MAX_N], st[MAX_N], ed[MAX_N], bb[MAX_N],
    seg2[MAX_N * 4];
bool check[MAX_N];
vector<int> v[MAX_N];
queue<int> q;
void DFS(int x) {
  check[x] = true;
  ans++;
  st[x] = ans;
  for (int i = 0; i < (int)v[x].size(); i++)
    if (check[v[x][i]] == false) {
      dist[v[x][i]] = dist[x] + 1;
      tab = max(tab, dist[v[x][i]]);
      DFS(v[x][i]);
    }
  ed[x] = ans;
}
void update(int l, int r, int val, int v = 1, int s = 1, int e = n + 1) {
  if (l <= s && e <= r + 1) {
    seg[v] += val;
    return;
  }
  int mid = (e + s) / 2;
  if (r < mid)
    update(l, r, val, v * 2, s, mid);
  else if (l >= mid)
    update(l, r, val, v * 2 + 1, mid, e);
  else {
    update(l, r, val, v * 2, s, mid);
    update(l, r, val, 2 * v + 1, mid, e);
  }
}
void update2(int l, int r, int val, int v = 1, int s = 0, int e = tab + 1) {
  if (l <= s && e <= r + 1) {
    seg2[v] += val;
    return;
  }
  int mid = (e + s) / 2;
  if (r < mid)
    update2(l, r, val, v * 2, s, mid);
  else if (l >= mid)
    update2(l, r, val, v * 2 + 1, mid, e);
  else {
    update2(l, r, val, v * 2, s, mid);
    update2(l, r, val, 2 * v + 1, mid, e);
  }
}
long long get(int x, int v = 1, int s = 1, int e = n + 1) {
  if (e - s <= 1) return seg[v];
  int mid = (e + s) / 2;
  if (x < mid)
    return seg[v] + get(x, v * 2, s, mid);
  else
    return seg[v] + get(x, v * 2 + 1, mid, e);
}
long long get2(int x, int v = 1, int s = 0, int e = tab + 1) {
  if (e - s <= 1) return seg2[v];
  int mid = (e + s) / 2;
  if (x < mid)
    return seg2[v] + get2(x, v * 2, s, mid);
  else
    return seg2[v] + get2(x, v * 2 + 1, mid, e);
}
int main() {
  int x, y;
  ios::sync_with_stdio(false);
  cin >> n >> qq;
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y;
    x--;
    y--;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  DFS(0);
  for (int i = 0; i < qq; i++) {
    long long u, val, d;
    cin >> x;
    if (x == 0) {
      cin >> u >> val >> d;
      u--;
      if (u == 0) {
        update2(0, min(d, tab), val);
      } else {
        long long up = dist[u];
        update(st[u], min((st[u] + d), ed[u]), val);
        if (up > d)
          update(st[u] - d, st[u] - 1, val);
        else if (up == d) {
          if (st[u] - dist[u] + 1 <= st[u] - 1)
            update(st[u] - dist[u] + 1, st[u] - 1, val);
          update(1, 1, val);
        } else {
          update2(0, d - dist[u], val);
          if (d - dist[u] < dist[u] &&
              st[u] - (2 * dist[u] - d - 1) <= st[u] - 1)
            update(st[u] - (2 * dist[u] - d - 1), st[u] - 1, val);
          else if (d - dist[u] == dist[u])
            update(st[u], st[u], -val);
          else if (d - dist[u] > dist[u] && st[u] <= st[u] - (2 * dist[u] - d))
            update(st[u], min(ed[u], st[u] - (2 * dist[u] - d)), -val);
        }
      }
    } else {
      cin >> u;
      u--;
      cout << get(st[u]) + get2(dist[u]) << endl;
    }
  }
  return 0;
}
