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;
struct FastIO {
  inline FastIO& operator>>(int& x) {
    x = 0;
    char f = 0, ch = getchar();
    while (ch > '9' || ch < '0') f |= (ch == '-'), ch = getchar();
    while (ch <= '9' && ch >= '0') x = x * 10 + ch - 48, ch = getchar();
    return x = (f ? -x : x), *this;
  }
  inline FastIO& operator>>(long long& x) {
    x = 0;
    char f = 0, ch = getchar();
    while (ch > '9' || ch < '0') f |= (ch == '-'), ch = getchar();
    while (ch <= '9' && ch >= '0') x = x * 10 + ch - 48, ch = getchar();
    return x = (f ? -x : x), *this;
  }
  inline FastIO& operator>>(double& x) {
    x = 0;
    char f = 0, ch = getchar();
    double d = 0.1;
    while (ch > '9' || ch < '0') f |= (ch == '-'), ch = getchar();
    while (ch <= '9' && ch >= '0') x = x * 10 + ch - 48, ch = getchar();
    if (ch == '.') {
      ch = getchar();
      while (ch <= '9' && ch >= '0')
        x += d * (ch ^ 48), d *= 0.1, ch = getchar();
    }
    return x = (f ? -x : x), *this;
  }
} read;
void file() {}
unsigned int log2(int x) {
  unsigned int ret;
  __asm__ __volatile__("bsrl %1, %%eax" : "=a"(ret) : "m"(x));
  return ret;
}
inline long long mul(long long a, long long b) {
  long long d = (long long)(a * (double)b / (int)(1e9 + 7) + 0.5);
  long long ret = a * b - d * (int)(1e9 + 7);
  if (ret < 0) ret += (int)(1e9 + 7);
  return ret;
}
int deg[200020];
struct hh {
  int t, nxt;
} edge[200020 << 1];
int head[200020], ecnt;
void make_edge(int f, int t) {
  edge[++ecnt] = (hh){t, head[f]};
  head[f] = ecnt;
  edge[++ecnt] = (hh){f, head[t]};
  head[t] = ecnt;
  deg[f]++;
  deg[t]++;
}
int n, K;
int dep[200020];
bool in[200020];
int vis[200020][25];
int main() {
  file();
  int i, j, k, x, y, z, sum = 0;
  read >> n >> K;
  if (n < 4) return puts("No"), 0;
  for (i = 1; i < n; i++) read >> x >> y, make_edge(x, y);
  for (i = 1; i <= n; i++)
    if (deg[i] == 1) {
      deg[i] = 0;
      vis[i][1] = 1;
      ++sum;
    }
  for (i = 2; i <= K + 1; i++) {
    for (j = 1; j <= n; j++)
      if (vis[j][i - 1]) {
        for (int k = head[j]; k; k = edge[k].nxt)
          if (!vis[edge[k].t][i - 2]) {
            if (vis[edge[k].t][i - 1]) return puts("No"), 0;
            vis[edge[k].t][i]++;
          }
      }
    for (j = 1; j <= n; j++)
      if (vis[j][i] == 1 || vis[j][i] == 2)
        return puts("No"), 0;
      else if (vis[j][i])
        ++sum;
    if (sum == n && i != K + 1) return puts("No"), 0;
  }
  if (sum != n) return puts("No"), 0;
  x = 0;
  for (i = 1; i <= n; i++)
    if (vis[i][K + 1]) ++x;
  if (x != 1) return puts("No"), 0;
  puts("Yes");
}
