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 = 100100;
struct segtree {
  int siz;
  vector<int> t;
  segtree() { siz = 0; }
  segtree(int k) {
    siz = 1;
    while (siz <= 2 * k + 1) {
      siz *= 2;
    }
    for (int i = 0; i < (siz); ++i) {
      t.push_back(0);
    }
    siz /= 2;
    for (int i = 0; i < ((int)t.size()); ++i) {
      t[i] = 0;
    }
  }
  void wloz(int a, int b, int c) {
    b = min(b, siz - 1);
    assert(b >= a);
    int l = a + siz;
    int p = b + siz;
    t[l] += c;
    if (p != l) t[p] += c;
    while (l / 2 != p / 2) {
      if (l % 2 == 0) {
        t[l + 1] += c;
      }
      if (p % 2 == 1) {
        t[p - 1] += c;
      }
      l /= 2;
      p /= 2;
    }
  }
  int daj(int v) {
    int res = 0;
    v += siz;
    while (v > 0) {
      res += t[v];
      v /= 2;
    }
    return res;
  }
};
vector<int> sas[MAXN];
int rzad[MAXN];
int kolumna[MAXN];
int ojc[MAXN];
int n, q;
void edgeIn(int a, int b) {
  sas[a].push_back(b);
  sas[b].push_back(a);
}
void readData() {
  scanf("%d", &n);
  scanf("%d", &q);
  int a, b;
  for (int i = 0; i < (n - 1); ++i) {
    scanf("%d%d", &a, &b);
    edgeIn(a, b);
  }
}
void bfs() {
  int start = 0;
  vector<int> bf;
  bf.push_back(1);
  ojc[1] = 1;
  for (int i = 0; i < ((int)bf.size()); ++i) {
    int obe = bf[i];
    int nei;
    for (int j = 0; j < ((int)sas[obe].size()); ++j) {
      nei = sas[obe][j];
      if (obe == 1) {
        kolumna[nei] = start;
        start++;
        rzad[nei] = 1;
        ojc[nei] = 1;
        bf.push_back(nei);
      } else if (ojc[nei] == 0) {
        ojc[nei] = obe;
        rzad[nei] = rzad[obe] + 1;
        kolumna[nei] = kolumna[obe];
        bf.push_back(nei);
      }
    }
  }
}
segtree tree[MAXN];
segtree general(MAXN);
void dodaj(int v, int x, int d) {
  int h = d - rzad[v];
  int kol = kolumna[v];
  if (h >= 0) {
    general.wloz(0, h, x);
    if (h + 1 <= tree[kol].siz - 2 && v != 1) {
      assert(rzad[v] + d >= h + 1);
      tree[kol].wloz(h + 1, rzad[v] + d, x);
    }
  } else {
    tree[kol].wloz(rzad[v] - d, rzad[v] + d, x);
  }
}
int wez(int v) { return general.daj(rzad[v]) + tree[kolumna[v]].daj(rzad[v]); }
int main() {
  readData();
  bfs();
  for (int i = 2; i <= (n); ++i) {
    if ((int)sas[i].size() == 1) {
      tree[kolumna[i]] = segtree(rzad[i] + 1);
    }
  }
  int sq, v, x, d;
  for (int i = 0; i < (q); ++i) {
    scanf("%d", &sq);
    if (sq == 0) {
      scanf("%d%d%d", &v, &x, &d);
      dodaj(v, x, d);
    } else {
      scanf("%d", &v);
      printf("%d\n", wez(v));
    }
  }
}
