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 M = 2e5 + 10;
const int N = 1e5 + 10;
struct apple {
  int be, to;
} edge[M];
int t[N], in[N], vist[N], zj[N], stak[N], dep[N];
int kk;
void addedge(int x, int y) {
  edge[++kk].to = y;
  edge[kk].be = t[x];
  t[x] = kk;
  in[y]++;
}
void dfs(int x) {
  int p, xx;
  vist[x] = 1;
  p = t[x];
  while (p) {
    xx = edge[p].to;
    if (!vist[xx]) {
      dep[xx] = dep[x] + 1;
      dfs(xx);
    }
    p = edge[p].be;
  }
}
void dfs2(int x, int r, int step) {
  int p, xx, i;
  vist[x] = 2;
  stak[step] = x;
  if (x == r) {
    for (i = 1; i <= step; i++) zj[i] = stak[i];
    return;
  }
  p = t[x];
  while (p) {
    xx = edge[p].to;
    if (vist[xx] == 1) {
      dfs2(xx, r, step + 1);
    }
    p = edge[p].be;
  }
}
void dfs3(int x) {
  int p, xx;
  vist[x] = 3;
  p = t[x];
  while (p) {
    xx = edge[p].to;
    if (vist[xx] == 2) {
      dep[xx] = dep[x] + 1;
      dfs3(xx);
    }
    p = edge[p].be;
  }
}
int main() {
  int n, k, x, y, l, r, root, zz, i;
  scanf("%d%d", &n, &k);
  for (i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    addedge(x, y);
    addedge(y, x);
  }
  for (l = 1; l <= n; l++)
    if (in[l] == 1) break;
  dfs(l);
  r = 0;
  for (i = 1; i <= n; i++) {
    if (dep[i] > dep[r]) r = i;
  }
  if (dep[r] % 2 == 1 || dep[r] / 2 != k) {
    printf("No");
    return 0;
  }
  dfs2(l, r, 1);
  zz = dep[r] / 2;
  root = zj[zz + 1];
  dep[root] = 0;
  dfs3(root);
  for (i = 1; i <= n; i++) {
    if ((in[i] == 1 && dep[i] == zz) ||
        (dep[i] < zz && i != root && in[i] >= 4) || (i == root && in[i] >= 3))
      continue;
    else {
      printf("No");
      return 0;
    }
  }
  printf("Yes");
  return 0;
}
