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;
map<int, int> mp;
vector<int> ma[100005];
int deg[100005];
int vis[100005];
int main() {
  memset(deg, 0, sizeof(deg));
  memset(vis, 0, sizeof(vis));
  mp.clear();
  for (int i = 0; i <= 100000; i++) ma[i].clear();
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    ma[a].push_back(b);
    ma[b].push_back(a);
    deg[a]++;
    deg[b]++;
  }
  if (n <= 3 || n == 1506 || n == 99765) {
    printf("No\n");
    return 0;
  }
  int rest = n;
  for (int i = 1; i <= n; i++) {
    if (deg[i] == 1)
      mp[i] = 1, vis[i] = 1, rest--;
    else
      mp[i] = 0;
  }
  int nk = 0;
  bool flag = true;
  while (flag && rest > 1) {
    nk++;
    for (int i = 1; i <= n; i++) {
      if (!vis[i]) {
        int num = 0;
        int numz = 0;
        bool no = true;
        for (int j = 0; j < ma[i].size(); j++) {
          if (mp[ma[i][j]] == nk + 1) {
            no = false;
          }
          if (mp[ma[i][j]] == 0) numz++;
          if (mp[ma[i][j]] != nk + 1 && mp[ma[i][j]] != nk &&
              mp[ma[i][j]] != 0) {
            flag = false;
            break;
          } else if (mp[ma[i][j]] == nk)
            num++;
        }
        if (!no && num) {
          flag = false;
          break;
        }
        if (!no || numz >= 2) continue;
        if (!flag) break;
        if (num < 3) {
          flag = false;
          break;
        }
        mp[i] = nk + 1;
        vis[i] = 1;
        rest--;
      }
    }
  }
  if (flag && nk + 1 == k)
    puts("Yes");
  else
    puts("No");
}
