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;
int depth[100005];
int idx_in_first[100005], pos[100005];
int idx_in_second[100005], indx[100005];
vector<vector<int>> tree, lazy, arr;
vector<int> adj[100005];
vector<pair<int, int>> V;
void dfs(int i, int node, int par, int dep) {
  arr[i].push_back(node);
  idx_in_second[node] = ((int)arr[i].size()) - 1;
  depth[node] = dep;
  pos[node] = i;
  V.push_back(make_pair(depth[node], node));
  for (int child : adj[node]) {
    if (child == par) continue;
    dfs(i, child, node, dep + 1);
  }
}
inline void propagate(int i, int node, int L, int R) {
  tree[i][node] += lazy[i][node];
  if (L != R) {
    lazy[i][2 * node] += lazy[i][node];
    lazy[i][2 * node + 1] += lazy[i][node];
  }
  lazy[i][node] = 0;
}
void update(int i, int node, int L, int R, int a, int b, int val) {
  propagate(i, node, L, R);
  if (R < a || L > b || R < L) return;
  if (L >= a && R <= b) {
    tree[i][node] += val;
    if (L != R) {
      lazy[i][2 * node] += val;
      lazy[i][2 * node + 1] += val;
    }
    return;
  }
  int mid = (L + R) >> 1;
  update(i, 2 * node, L, mid, a, b, val);
  update(i, 2 * node + 1, mid + 1, R, a, b, val);
  tree[i][node] = tree[i][2 * node] + tree[i][2 * node + 1];
}
int query(int i, int node, int L, int R, int a, int b) {
  propagate(i, node, L, R);
  if (R < a || L > b || R < L) return 0;
  if (L >= a && R <= b) return tree[i][node];
  int mid = (L + R) >> 1;
  return query(i, 2 * node, L, mid, a, b) +
         query(i, 2 * node + 1, mid + 1, R, a, b);
}
int main() {
  int n, q, a, b, v, x, d, qs;
  scanf("%d%d", &n, &q);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  tree.resize(1 + ((int)adj[1].size()));
  lazy.resize(1 + ((int)adj[1].size()));
  arr.resize(1 + ((int)adj[1].size()));
  int idx = 1;
  for (int child : adj[1]) dfs(idx++, child, 1, 1);
  sort((V.begin()), (V.end()));
  for (int i = 0; i < ((int)V.size()); i++) {
    arr[0].push_back(V[i].second);
    indx[V[i].first]++;
    idx_in_first[V[i].second] = i;
  }
  for (int i = 1; i <= n; i++) indx[i] += indx[i - 1];
  tree[0].resize(4 * n);
  lazy[0].resize(4 * n);
  for (int i = 1; i < idx; i++) {
    tree[i].resize(4 * ((int)arr[i].size()));
    lazy[i].resize(4 * ((int)arr[i].size()));
  }
  int node1 = 0;
  while (q--) {
    scanf("%d", &qs);
    if (qs == 0) {
      scanf("%d%d%d", &v, &x, &d);
      int p = pos[v], i1 = idx_in_first[v], i2 = idx_in_second[v];
      if (v == 1) {
        node1 += x;
        update(0, 1, 0, n - 2, 0, indx[d] - 1, x);
        continue;
      }
      int dd = min(i2 + d, ((int)arr[p].size()) - 1);
      update(p, 1, 0, ((int)arr[p].size()) - 1, i2, dd, x);
      dd = max(0, i2 - d);
      update(p, 1, 0, ((int)arr[p].size()) - 1, dd, i2 - 1, x);
      if (i2 + 1 <= d) {
        dd = d - i2 - 1;
        node1 += x;
        update(0, 1, 0, n - 2, 0, indx[dd] - 1, x);
        update(p, 1, 0, ((int)arr[p].size()) - 1, 0, dd - 1, -x);
      }
    } else {
      scanf("%d", &v);
      if (v == 1) {
        printf("%d\n", node1);
        continue;
      }
      int ans = query(0, 1, 0, n - 2, idx_in_first[v], idx_in_first[v]);
      ans += query(pos[v], 1, 0, ((int)arr[pos[v]].size()) - 1,
                   idx_in_second[v], idx_in_second[v]);
      printf("%d\n", ans);
    }
  }
  return 0;
}
