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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N = 1e5 + 100;
const int M = 1 << 5;
const int LEN = 1e4 + 5;
const int hashMod = 1e6 + 5;
const int inf = 0x77777777;
const double eps = 1e-5;
const long long linf = 0x7777777777777777ll;
const double finf = 1e50;
const double pi = acos(-1.0);
const int mod = 1e9 + 7;
template <class T>
inline T sqr(T x) {
  return x * x;
}
vector<bool> vis;
vector<vector<int> > rel, BIT;
vector<int> depth;
vector<pair<int, int> > id;
void init(int n) {
  vector<int> tmp;
  tmp.clear();
  vis.clear();
  rel.clear();
  BIT.clear();
  depth.clear();
  id.clear();
  for (int i = 0; i < (n + 1); i++)
    vis.push_back(false), rel.push_back(tmp), BIT.push_back(tmp),
        depth.push_back(0), id.push_back(make_pair(0, 0));
}
void input(int n) {
  int x, y;
  for (int i = 0; i < (n - 1); i++) {
    scanf("%d%d", &x, &y);
    rel[x].push_back(y);
    rel[y].push_back(x);
  }
}
void build() {
  int maxDepth = 0;
  vis[1] = true;
  for (int i = 0; i < (((int)rel[1].size())); i++) {
    int cur = rel[1][i], curDepth = 1;
    BIT[i + 1].push_back(0);
    while (true) {
      if (vis[cur]) break;
      vis[cur] = true;
      BIT[i + 1].push_back(0);
      id[cur] = make_pair(i + 1, curDepth);
      depth[cur] = curDepth++;
      for (int j = 0; j < (((int)rel[cur].size())); j++) {
        int t = rel[cur][j];
        if (vis[t]) continue;
        cur = t;
      }
    }
    maxDepth = max(maxDepth, curDepth);
  }
  for (int i = 0; i < (maxDepth); i++) BIT[0].push_back(0);
}
void add(int x, int y, int v) {
  for (int i = y; i >= 1; i -= i & (-i)) BIT[x][i] += v;
}
int getVal(int x, int y) {
  int ret = 0;
  for (int i = y; i < ((int)BIT[x].size()); i += i & (-i)) ret += BIT[x][i];
  return ret;
}
int root;
int query(int x) {
  int fi = id[x].first, se = id[x].second;
  if (!fi || !se) return root;
  return getVal(fi, se) + getVal(0, se);
}
void insert(int v, int x, int d) {
  int dep = depth[v], fi = id[v].first, se = id[v].second;
  if (dep == d) {
    root += x;
  } else if (dep > d) {
    add(fi, dep - d - 1, -x);
  } else {
    root += x;
    add(0, min(d - dep, ((int)BIT[0].size()) - 1), x);
    add(fi, min(d - dep, ((int)BIT[fi].size()) - 1), -x);
  }
  add(fi, min(se + d, ((int)BIT[fi].size()) - 1), x);
}
int main() {
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    int op, v, x, d;
    init(n);
    input(n);
    build();
    root = 0;
    for (int i = 0; i < (m); i++) {
      scanf("%d", &op);
      if (op) {
        scanf("%d", &x);
        printf("%d\n", query(x));
      } else {
        scanf("%d%d%d", &v, &x, &d);
        insert(v, x, d);
      }
    }
  }
  return 0;
}
