Someone give a strange birthday present to Ivan. It is hedgehog — connected undirected graph in which one vertex has degree at least 3 (we will call it center) and all other vertices has degree 1. Ivan thought that hedgehog is too boring and decided to make himself k-multihedgehog.

Let us define k-multihedgehog as follows:

  * 1-multihedgehog is hedgehog: it has one vertex of degree at least 3 and some vertices of degree 1.
  * For all k ≥ 2, k-multihedgehog is (k-1)-multihedgehog in which the following changes has been made for each vertex v with degree 1: let u be its only neighbor; remove vertex v, create a new hedgehog with center at vertex w and connect vertices u and w with an edge. New hedgehogs can differ from each other and the initial gift. 



Thereby k-multihedgehog is a tree. Ivan made k-multihedgehog but he is not sure that he did not make any mistakes. That is why he asked you to check if his tree is indeed k-multihedgehog.

Input

First line of input contains 2 integers n, k (1 ≤ n ≤ 10^{5}, 1 ≤ k ≤ 10^{9}) — number of vertices and hedgehog parameter.

Next n-1 lines contains two integers u v (1 ≤ u,    v ≤ n;    u ≠ v) — indices of vertices connected by edge.

It is guaranteed that given graph is a tree.

Output

Print "Yes" (without quotes), if given graph is k-multihedgehog, and "No" (without quotes) otherwise.

Examples

Input

14 2
1 4
2 4
3 4
4 13
10 5
11 5
12 5
14 5
5 13
6 7
8 6
13 6
9 6


Output

Yes


Input

3 1
1 3
2 3


Output

No

Note

2-multihedgehog from the first example looks like this:

<image>

Its center is vertex 13. Hedgehogs created on last step are: [4 (center), 1, 2, 3], [6 (center), 7, 8, 9], [5 (center), 10, 11, 12, 13].

Tree from second example is not a hedgehog because degree of center should be at least 3.
#include <bits/stdc++.h>
using namespace std;
struct node {
  int val;
  long long d;
};
map<int, set<int> > hed;
int vis[100010];
vector<node> deg;
long long fat_deg[100010];
int main() {
  int n;
  long long k;
  k = 0;
  while (cin >> n >> k) {
    if (n == 1) {
      cout << "No" << endl;
      continue;
    }
    int x, y;
    hed.clear();
    deg.clear();
    memset(vis, 0, sizeof(vis));
    for (int i = 0; i < 100010; i++) {
      fat_deg[i] = 1e9 + 7;
    }
    for (int i = 0; i < n - 1; i++) {
      cin >> x >> y;
      vis[x]++;
      vis[y]++;
      hed[x].insert(y);
      hed[y].insert(x);
    }
    node tmp;
    for (int i = 1; i <= n; i++) {
      if (vis[i] == 1) {
        tmp.d = 1;
        tmp.val = i;
        deg.push_back(tmp);
      }
    }
    int son;
    long long d;
    int flag1 = 0;
    int fat;
    for (int i = 0; i < deg.size(); i++) {
      int flag = 0;
      son = deg[i].val;
      d = deg[i].d;
      for (set<int>::iterator it = hed[son].begin(); it != hed[son].end();
           it++) {
        fat = *it;
        if ((vis[fat] <= 3 && fat_deg[fat] != 1 && n - deg.size() > 1) ||
            (vis[fat] < 3 && fat_deg[fat] != 1 && n - deg.size() <= 1)) {
          flag = 1;
          break;
        }
        hed[fat].erase(son);
        if (fat_deg[fat] != 1e9 + 7) {
          if (fat_deg[fat] != d + 1) {
            flag = 1;
            break;
          }
        } else {
          fat_deg[fat] = d + 1;
        }
        if (hed[fat].size() <= 1) {
          tmp.d = fat_deg[fat];
          tmp.val = fat;
          deg.push_back(tmp);
        }
      }
      hed.erase(son);
      if (flag != 0) {
        flag1 = flag;
        break;
      }
    }
    if (flag1 == 1) {
      cout << "No" << endl;
      continue;
    } else if (flag1 == 0 && deg[deg.size() - 1].d - 1 == k) {
      cout << "Yes" << endl;
    } else {
      cout << "No" << endl;
    }
  }
  return 0;
}
