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 = 100050;
int len[maxn], id[maxn], depth[maxn], n, q;
vector<int> sum[maxn], lazy[maxn], tree[maxn];
void dfs(int f, int u, int dp, int no);
void update(int idx, int L, int R, int add, int l, int r, int rt);
int query(int idx, int L, int R, int l, int r, int rt);
void push_down(int idx, int rt, int m);
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    tree[u].push_back(v);
    tree[v].push_back(u);
  }
  dfs(1, 1, 1, 0);
  int nn = tree[1].size() + 1;
  for (int i = 1; i <= nn; i++) len[0] = max(len[i], len[0]);
  for (int i = 0; i < nn; i++)
    for (int j = 0; j < len[i] * 5; j++) {
      sum[i].push_back(0);
      lazy[i].push_back(0);
    }
  while (q--) {
    int op, v, x, d;
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d", &v);
      int ans = 0;
      if (v != 1) ans += query(id[v], depth[v], depth[v], 1, len[id[v]], 1);
      ans += query(0, depth[v], depth[v], 1, len[0], 1);
      printf("%d\n", ans);
    } else {
      scanf("%d%d%d", &v, &d, &x);
      if (depth[v] - 1 > x)
        update(id[v], depth[v] - x, min(len[id[v]], depth[v] + x), d, 1,
               len[id[v]], 1);
      else {
        int L = x - (depth[v] - 1);
        if (2 + L <= len[id[v]] && v != 1)
          update(id[v], 2 + L, min(len[id[v]], depth[v] + x), d, 1, len[id[v]],
                 1);
        update(0, 1, min(len[0], 1 + L), d, 1, len[0], 1);
      }
    }
  }
  return 0;
}
void dfs(int f, int u, int dp, int no) {
  depth[u] = dp;
  id[u] = no;
  len[no] = dp;
  for (int i = 0; i < tree[u].size(); i++) {
    int v = tree[u][i];
    if (v == f) continue;
    if (u == 1) no++;
    dfs(u, v, dp + 1, no);
  }
}
void push_down(int idx, int rt, int m) {
  if (lazy[idx][rt]) {
    lazy[idx][rt << 1] += lazy[idx][rt];
    lazy[idx][rt << 1 | 1] += lazy[idx][rt];
    sum[idx][rt << 1] += lazy[idx][rt] * (m - (m >> 1));
    sum[idx][rt << 1 | 1] += lazy[idx][rt] * (m >> 1);
    lazy[idx][rt] = 0;
  }
}
int query(int idx, int L, int R, int l, int r, int rt) {
  if (L <= l && r <= R) return sum[idx][rt];
  int m = (l + r) >> 1;
  push_down(idx, rt, r - l + 1);
  int ret = 0;
  if (L <= m) ret += query(idx, L, R, l, m, rt << 1);
  if (R > m) ret += query(idx, L, R, m + 1, r, rt << 1 | 1);
  return ret;
}
void update(int idx, int L, int R, int add, int l, int r, int rt) {
  if (L <= l && r <= R) {
    sum[idx][rt] += add * (r - l + 1);
    lazy[idx][rt] += add;
    return;
  }
  int m = (l + r) >> 1;
  push_down(idx, rt, r - l + 1);
  if (L <= m) update(idx, L, R, add, l, m, rt << 1);
  if (R > m) update(idx, L, R, add, m + 1, r, rt << 1 | 1);
  sum[idx][rt] = sum[idx][rt << 1] + sum[idx][rt << 1 | 1];
}
