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;
int n, k, x, y, ans = 1, der[1000005], mx[1000005], bas;
vector<int> g[1000005];
void dfs(int node, int par, int der) {
  int sz = g[node].size();
  if (sz == 1) {
    if (der != k) {
      ans = 0;
    }
    return;
  }
  if (par and sz > 1 and sz <= 3) {
    ans = 0;
    return;
  }
  if (!par and sz == 2) {
    ans = 0;
    return;
  }
  for (int i = 0; i < sz; i++)
    if (g[node][i] != par) dfs(g[node][i], node, der + 1);
}
int hazirla(int node, int par, int dr) {
  mx[node] = der[node] = dr;
  int xx = 0;
  for (int i = 0; i < g[node].size(); i++)
    if (g[node][i] != par) {
      mx[node] = max(mx[node], hazirla(g[node][i], node, dr + 1));
      xx = 1;
    }
  if (!xx) bas = node;
  return mx[node];
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  if (n <= 3) {
    puts("No");
    return 0;
  }
  hazirla(1, 0, 0);
  int ebas = bas;
  hazirla(bas, 0, 0);
  int node = ebas, par = 0, f = 1, en = 0;
  while (f) {
    f = 0;
    for (int i = 0; i < g[node].size(); i++)
      if (g[node][i] != par) {
        if (mx[g[node][i]] - der[node] > en) {
          for (int j = 0; j < g[node].size(); j++)
            if (g[node][j] != par and g[node][j] != g[node][i])
              en = max(en, mx[g[node][j]] - der[node] + 1);
          en = max(en, der[g[node][i]]);
          par = node;
          node = g[node][i];
          f = 1;
          break;
        }
      }
  }
  dfs(node, 0, 0);
  if (ans)
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}
