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 N1 = 200005;
const int inf = 0x3f3f3f3f;
struct Edge {
  int to[N1], nxt[N1], head[N1], cte;
  void ae(int u, int v) {
    cte++;
    to[cte] = v;
    nxt[cte] = head[u];
    head[u] = cte;
  }
} e;
int n, K;
int deg[N1], b[N1], h[N1];
int vis[N1], que[N1], hd, tl;
int main() {
  scanf("%d%d", &n, &K);
  int x, y;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    e.ae(x, y), e.ae(y, x);
    deg[x]++, deg[y]++;
  }
  if (K > n || n < 4) {
    puts("No");
    return 0;
  }
  int v;
  hd = 1;
  tl = 0;
  for (int i = 1; i <= n; i++) {
    b[i] = deg[i];
    if (deg[i] != 1) continue;
    h[i] = 1;
    que[++tl] = i;
  }
  while (hd <= tl) {
    x = que[hd];
    hd++;
    vis[x] = 1;
    for (int j = e.head[x]; j; j = e.nxt[j]) {
      v = e.to[j];
      if (vis[v]) continue;
      deg[v]--;
      if (!h[v]) {
        h[v] = h[x] + 1;
      } else if (h[v] != h[x] + 1) {
        puts("No");
        return 0;
      }
      if (deg[v] == 1) que[++tl] = v;
    }
  }
  int root = 0, mh = 0;
  for (int i = 1; i <= n; i++) {
    if (h[i] > mh) mh = h[i], root = i;
  }
  if (b[root] < 3 || deg[root] != 0 || mh != K + 1) {
    puts("No");
    return 0;
  }
  for (int i = 1; i <= n; i++)
    if (b[i] != 1 && i != root) {
      if (b[i] < 4) {
        puts("No");
        return 0;
      }
    }
  puts("Yes");
  return 0;
}
