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;
struct node {
  int v, x, d;
  node(int v = 0, int x = 0, int d = 0) : v(v), x(x), d(d) {}
};
int n, q;
vector<int> G[200005];
int in[200005], out[200005], depth[200005], subtree[200005];
int arr[200005], where[200005];
int suff[200005], remdep[200005];
int par[200005];
int a[200005];
int first[200005];
node queries[2048 + 1];
int cnt;
int TIME, idx;
void dfs(int u, int p) {
  in[u] = TIME++;
  subtree[u] = 1;
  par[u] = p;
  for (int i = 0; i < (int)G[u].size(); i++) {
    int v = G[u][i];
    if (v == p) continue;
    depth[v] = depth[u] + 1;
    if (u == 1)
      first[v] = v;
    else
      first[v] = first[u];
    dfs(v, u);
    subtree[u] += subtree[v];
  }
  arr[idx] = u;
  where[u] = idx;
  idx++;
  out[u] = TIME - 1;
}
int dist(int u, int v) {
  if (depth[u] > depth[v]) swap(u, v);
  if (in[u] <= in[v] && out[u] >= out[v]) {
    return depth[v] - depth[u];
  }
  return depth[u] + depth[v];
}
int get(int u, int by) {
  if (by >= 0) {
    if (subtree[u] - 1 < by) return -1;
    return arr[where[u] - by];
  } else {
    by = -by;
    if (depth[u] - 1 < by) return -1;
    return arr[where[u] + by];
  }
}
void bfs() {
  memset(suff, 0, sizeof suff);
  memset(remdep, 0, sizeof remdep);
  int al = 0;
  node tmp;
  for (int i = 0; i < cnt; i++) {
    tmp = queries[i];
    if (tmp.v != 1) {
      int y = get(tmp.v, -tmp.d);
      if (y == -1) y = first[tmp.v];
      suff[y] += tmp.x;
      if (get(tmp.v, tmp.d + 1) != -1) suff[get(tmp.v, +tmp.d + 1)] -= tmp.x;
      int rem = tmp.d - depth[tmp.v];
      if (rem >= 0) {
        al += tmp.x;
        remdep[rem + 1] -= tmp.x;
        suff[first[tmp.v]] -= tmp.x;
        if (get(first[tmp.v], rem) != -1) suff[get(first[tmp.v], rem)] += tmp.x;
      }
    } else {
      al += tmp.x;
      remdep[tmp.d + 1] -= tmp.x;
    }
  }
  queue<int> q;
  q.push(1);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    suff[u] += remdep[depth[u]];
    a[u] += al + suff[u];
    for (int i = 0; i < (int)G[u].size(); i++) {
      int v = G[u][i];
      if (v == par[u]) continue;
      suff[v] += suff[u];
      q.push(v);
    }
  }
  cnt = 0;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    G[x].push_back(y), G[y].push_back(x);
  }
  dfs(1, -1);
  int type, v, x, d;
  while (q--) {
    scanf("%d", &type);
    if (type == 0) {
      scanf("%d%d%d", &v, &x, &d);
      queries[cnt++] = (node(v, x, d));
      if (cnt == 2048) bfs();
    } else {
      scanf("%d", &v);
      int ans = a[v];
      for (int i = 0; i < cnt; i++) {
        if (dist(v, queries[i].v) <= queries[i].d) ans += queries[i].x;
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}
