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>
#pragma GCC optimize("Ofast")
using namespace std;
template <class T>
T Gcd(T x, T y) {
  return y ? Gcd(y, x % y) : x;
}
template <class T>
T Lcm(T x, T y) {
  return x / Gcd(x, y) * y;
}
template <class T>
void Adj(T &x, T y) {
  if (x >= y) x %= y;
  while (x < 0) x += y;
}
const long long MOD = 998244353LL;
const int M = 205;
const int N = 100005;
int h[N], n, m, dep[N], f[N], fa[N], tot, op, ed, t, tt, k, d[N], cnt[N];
struct u {
  int nx, to;
} g[N << 1];
inline void ade(int x, int y) {
  g[++tot] = {h[x], y};
  h[x] = tot;
}
inline void bfs(int x) {
  memset((f), (0), sizeof((f)));
  d[1] = x;
  f[x] = 1;
  dep[x] = 0;
  for (op = ed = 1; op <= ed; op++) {
    int now = d[op];
    for (int i = h[now]; i; i = g[i].nx)
      if (!f[g[i].to]) {
        d[++ed] = g[i].to;
        f[d[ed]] = 1;
        fa[d[ed]] = now;
        dep[d[ed]] = dep[now] + 1;
      }
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &t, &tt);
    ade(t, tt);
    ade(tt, t);
    cnt[tt]++;
    cnt[t]++;
  }
  bfs(1);
  bfs(d[ed]);
  if (dep[d[ed]] & 1) {
    puts("No");
    exit(0);
  }
  k = dep[d[ed]] >> 1;
  t = d[ed];
  while (k--) t = fa[t];
  if (cnt[t] < 3) {
    puts("No");
    exit(0);
  }
  bfs(t);
  for (int i = 1; i <= n; i++)
    if (i != t) {
      if (dep[i] == m && cnt[i] != 1) {
        puts("No");
        exit(0);
      }
      if (dep[i] != m && cnt[i] <= 3) {
        puts("No");
        exit(0);
      }
    }
  puts("Yes");
  return 0;
}
