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 = 211111;
const double eps = 1e-8;
const double pi = acos(-1.0);
int order[maxn];
int dep[maxn];
int E, head[maxn];
int h[maxn];
pair<int, int> p[maxn];
int lhs[maxn], rhs[maxn];
struct node {
  int v, next;
  node() {}
  node(int v, int next) : v(v), next(next) {}
} edge[maxn << 2];
void add_edge(int u, int v) {
  edge[E] = node(v, head[u]);
  head[u] = E++;
}
int cnt;
int deep[maxn];
void dfs(int u, int pre = -1) {
  order[u] = ++cnt;
  if (~pre) deep[u] = dep[u];
  int i, v;
  for (i = head[u]; ~i; i = edge[i].next) {
    v = edge[i].v;
    if (v == pre) continue;
    dep[v] = dep[u] + 1;
    dfs(v, u);
    deep[u] = max(deep[v], deep[u]);
  }
}
template <class T>
class SegTree {
 public:
  T add[maxn << 2];
  void pushdown(int rt) {
    if (add[rt]) {
      add[rt << 1] += add[rt];
      add[rt << 1 | 1] += add[rt];
      add[rt] = 0;
    }
  }
  void update(int L, int R, T x, int l, int r, int rt) {
    if (l >= L && r <= R) {
      add[rt] += x;
      return;
    }
    pushdown(rt);
    int m = (l + r) >> 1;
    if (m >= L) update(L, R, x, l, m, rt << 1);
    if (m < R) update(L, R, x, m + 1, r, rt << 1 | 1);
  }
  T query(int p, int l, int r, int rt) {
    if (l == r) {
      return add[rt];
    }
    pushdown(rt);
    int m = (l + r) >> 1;
    T t;
    if (p <= m)
      t = query(p, l, m, rt << 1);
    else
      t = query(p, m + 1, r, rt << 1 | 1);
    return t;
  }
};
SegTree<long long> Order, Dep;
int main() {
  ios::sync_with_stdio(false);
  memset(head, -1, sizeof(head));
  E = 0;
  int i, u, v, n, m;
  cin >> n >> m;
  for (i = 1; i < n; ++i) {
    cin >> u >> v;
    add_edge(u, v);
    add_edge(v, u);
  }
  dep[1] = 1;
  dfs(1);
  memset(lhs, 0x7f, sizeof(lhs));
  memset(rhs, 0, sizeof(rhs));
  for (i = 1; i <= n; ++i) p[i] = pair<int, int>(dep[i], i);
  sort(p + 1, p + n + 1);
  for (i = 1; i <= n; ++i) h[p[i].second] = i;
  for (i = 1; i <= n; ++i) {
    rhs[p[i].first] = max(rhs[p[i].first], i);
  }
  for (i = 2; i <= deep[v]; ++i) {
    rhs[i] = max(rhs[i], rhs[i - 1]);
  }
  long long x;
  int d, tmp, tip, a, b, l, r, k;
  while (m--) {
    cin >> tip;
    if (!tip) {
      cin >> v >> x >> d;
      k = order[v];
      l = order[v] - d;
      if (dep[v] + d <= deep[v])
        r = k + d;
      else
        r = k + deep[v] - dep[v];
      if (v == 1) {
        r = d + 1;
        r = min(p[n].first, r);
        r = rhs[r];
        Dep.update(1, r, x, 1, n, 1);
        continue;
      }
      if (dep[v] - d > 1) {
        Order.update(l, r, x, 1, n, 1);
      } else if (dep[v] == d + 1) {
        Order.update(1, 1, x, 1, n, 1);
        tmp = d - dep[v] + 2;
        a = k - dep[v] + 2;
        Order.update(a, r, x, 1, n, 1);
      } else {
        tmp = d - dep[v] + 2;
        tmp = min(p[n].first, tmp);
        a = 1, b = rhs[tmp];
        Dep.update(a, b, x, 1, n, 1);
        a = k - dep[v] + 2, b = a + tmp - 2;
        b = min(b, r);
        Order.update(a, b, -x, 1, n, 1);
        Order.update(a, r, x, 1, n, 1);
      }
    } else {
      cin >> v;
      a = h[v], b = order[v];
      long long ans = Order.query(b, 1, n, 1) + Dep.query(a, 1, n, 1);
      cout << ans << endl;
    }
  }
  return 0;
}
