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 N = 1e5 + 10;
vector<int> g[N];
pair<int, int> chain[N];
int st1[4 * N], st2[4 * N];
int lazy1[4 * N], lazy2[4 * N];
int n, k = 1;
int depth[N], last[N];
int hes1[N], hes2[N];
int inv1[N], inv2[N];
bool mark[N];
int in[N];
int ch;
void dfs(int v, int prev, int d, int lanac) {
  hes1[v] = ++k;
  inv1[k] = v;
  depth[v] = d;
  in[v] = lanac;
  chain[lanac].second = v;
  for (auto u : g[v]) {
    if (u == prev) continue;
    dfs(u, v, d + 1, lanac);
  }
}
void bfs(int source) {
  queue<int> q;
  q.push(source);
  k = 0;
  last[0] = 1;
  while (!q.empty()) {
    int xt = q.front();
    q.pop();
    mark[xt] = true;
    hes2[xt] = ++k;
    inv2[k] = xt;
    last[depth[xt]] = max(hes2[xt], last[depth[xt]]);
    for (auto u : g[xt]) {
      if (!mark[u]) q.push(u);
    }
  }
}
void propagate1(int idx, int left, int right) {
  if (lazy1[idx] != 0) {
    st1[idx] += lazy1[idx] * (right - left + 1);
    if (left < right) {
      lazy1[2 * idx] += lazy1[idx];
      lazy1[2 * idx + 1] += lazy1[idx];
    }
    lazy1[idx] = 0;
  }
}
void propagate2(int idx, int left, int right) {
  if (lazy2[idx] != 0) {
    st2[idx] += lazy2[idx] * (right - left + 1);
    if (left < right) {
      lazy2[2 * idx] += lazy2[idx];
      lazy2[2 * idx + 1] += lazy2[idx];
    }
    lazy2[idx] = 0;
  }
}
void update1(int idx, int left, int right, int l, int r, int val) {
  propagate1(idx, left, right);
  if (left > right || left > r || right < l) return;
  if (l <= left && right <= r) {
    st1[idx] += val * (right - left + 1);
    if (left < right) {
      lazy1[2 * idx] += val;
      lazy1[2 * idx + 1] += val;
    }
    return;
  }
  update1(2 * idx, left, (left + right) / 2, l, r, val);
  update1(2 * idx + 1, (left + right) / 2 + 1, right, l, r, val);
  st1[idx] = st1[2 * idx] + st1[2 * idx + 1];
}
void update2(int idx, int left, int right, int l, int r, int val) {
  propagate2(idx, left, right);
  if (left > right || left > r || right < l) return;
  if (l <= left && right <= r) {
    st2[idx] += val * (right - left + 1);
    if (left < right) {
      lazy2[2 * idx] += val;
      lazy2[2 * idx + 1] += val;
    }
    return;
  }
  update2(2 * idx, left, (left + right) / 2, l, r, val);
  update2(2 * idx + 1, (left + right) / 2 + 1, right, l, r, val);
  st2[idx] = st2[2 * idx] + st2[2 * idx + 1];
}
int query1(int idx, int left, int right, int l, int r) {
  propagate1(idx, left, right);
  if (left > right || left > r || right < l) return 0;
  if (l <= left && right <= r) return st1[idx];
  int ret = 0;
  ret += query1(2 * idx, left, (left + right) / 2, l, r);
  ret += query1(2 * idx + 1, (left + right) / 2 + 1, right, l, r);
  return ret;
}
int query2(int idx, int left, int right, int l, int r) {
  propagate2(idx, left, right);
  if (left > right || left > r || right < l) return 0;
  if (l <= left && right <= r) return st2[idx];
  int ret = 0;
  ret += query2(2 * idx, left, (left + right) / 2, l, r);
  ret += query2(2 * idx + 1, (left + right) / 2 + 1, right, l, r);
  return ret;
}
int q;
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (auto xt : g[1]) {
    chain[++ch].first = xt;
    dfs(xt, 1, 1, ch);
  }
  bfs(1);
  for (int i = 1; i <= n; i++)
    if (last[i] == 0) last[i] = n;
  int type, node, x, d;
  while (q--) {
    scanf("%d%d", &type, &node);
    if (type == 1)
      printf("%d\n", query1(1, 1, n, hes1[node], hes1[node]) +
                         query2(1, 1, n, hes2[node], hes2[node]));
    else {
      scanf("%d%d", &x, &d);
      if (node == 1) {
        update2(1, 1, n, 1, last[d], x);
        continue;
      }
      if (d < depth[node])
        update1(1, 1, n, hes1[node] - d,
                min(hes1[node] + d, hes1[chain[in[node]].second]), x);
      else if (d == depth[node]) {
        update1(1, 1, n, hes1[node] - d + 1,
                min(hes1[node] + d, hes1[chain[in[node]].second]), x);
        update2(1, 1, n, 1, 1, x);
      } else {
        int more = d - depth[node];
        update2(1, 1, n, 1, last[more], x);
        update1(1, 1, n, hes1[chain[in[node]].first] + more,
                min(hes1[node] + d, hes1[chain[in[node]].second]), x);
      }
    }
  }
  return 0;
}
