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;
const int N = 100001;
vector<int> g[N];
int in[N], dep[N], P[N];
int n, k;
void dfs(int c, int p, int d) {
  dep[c] = d;
  P[c] = p;
  for (int i = 0; i < g[c].size(); i++) {
    int u = g[c][i];
    if (u != p) {
      dfs(u, c, d + 1);
    }
  }
}
int findCenter() {
  dfs(1, -1, 0);
  int x = 1;
  for (int i = 2; i < n + 1; i++)
    if (dep[i] > dep[x]) {
      x = i;
    }
  dfs(x, -1, 0);
  x = 1;
  for (int i = 2; i < n + 1; i++)
    if (dep[i] > dep[x]) {
      x = i;
    }
  if (dep[x] != 2 * k) return -1;
  for (int i = 0; i < k; i++) x = P[x];
  return x;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
    in[u]++;
    in[v]++;
  }
  if (n < 4) {
    printf("No\n");
    return 0;
  }
  int c = findCenter();
  if (c == -1) {
    printf("No\n");
    return 0;
  }
  dfs(c, -1, 0);
  for (int i = 1; i < n + 1; i++) {
    if (c == i) {
      if (in[i] < 3) {
        printf("No\n");
        return 0;
      }
    } else if (in[i] == 1) {
      if (dep[i] != k) {
        printf("No\n");
        return 0;
      }
    } else {
      if (in[i] < 4) {
        printf("No\n");
        return 0;
      }
    }
  }
  printf("Yes\n");
  return 0;
}
