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;
int N, Q;
struct bit {
  int *BIT;
  int N;
  bit() {}
  void init(int sz) {
    N = sz;
    BIT = new int[N + 2];
    for (int i = 0; i <= N; ++i) {
      BIT[i] = 0;
    }
  }
  int ask(int x) {
    ++x;
    int ret = 0;
    while (x) {
      ret += BIT[x];
      x -= x & -x;
    }
    return ret;
  }
  void inc(int x, int val) {
    ++x;
    while (x <= N) {
      BIT[x] += val;
      x += x & -x;
    }
  }
};
bit chainsums[MAXN];
bit rootsums;
vector<int> conn[MAXN];
int pos[MAXN];
int num[MAXN];
int cnts[MAXN];
void build(int cur, int prv, int curnum, int &curpos) {
  ++curpos;
  pos[cur] = curpos;
  num[cur] = curnum;
  for (typeof((conn[cur]).begin()) i = (conn[cur]).begin(),
                                   _i = (conn[cur]).end();
       i != _i; ++i) {
    if (*i != prv) {
      build(*i, cur, curnum, curpos);
    }
  }
}
void buildall() {
  int curnum = 0;
  for (typeof((conn[1]).begin()) i = (conn[1]).begin(), _i = (conn[1]).end();
       i != _i; ++i) {
    int cnt = 0;
    build(*i, 1, curnum, cnt);
    chainsums[curnum].init(cnt + 1);
    ++curnum;
  }
  rootsums.init(N);
}
int query(int x) {
  int ans = rootsums.ask(pos[x]);
  if (x != 1) {
    int part2 = chainsums[num[x]].ask(pos[x]);
    ans += part2;
  }
  return ans;
}
void add(int node, int val, int dst) {
  int rdst = dst - pos[node];
  if (rdst >= 0) {
    rootsums.inc(0, val);
    rootsums.inc(rdst + 1, -val);
  }
  bit &b = chainsums[num[node]];
  int top = pos[node] - dst;
  int low = pos[node] + dst + 1;
  if (top <= rdst) {
    top = rdst + 1;
  }
  if (top < low) {
    b.inc(top, val);
    b.inc(low, -val);
  }
}
int main() {
  scanf("%d %d", &N, &Q);
  for (int i = 1; i < N; ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    conn[a].push_back(b);
    conn[b].push_back(a);
  }
  buildall();
  for (int q = 0; q < Q; ++q) {
    int qtype;
    scanf("%d", &qtype);
    if (qtype == 0) {
      int a, b, c;
      scanf("%d %d %d", &a, &b, &c);
      add(a, b, c);
    } else if (qtype == 1) {
      int p;
      scanf("%d", &p);
      int ans = query(p);
      printf("%d\n", ans);
    }
  }
  return 0;
}
