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 mod = 998244353;
const int inf = 0x3f3f3f3f;
long long gcd(long long x, long long y) {
  if (y == 0)
    return x;
  else
    return gcd(y, x % y);
}
long long ksm(long long a, long long b) {
  long long r = 1;
  while (b) {
    if (b & 1) r = (r * a) % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return r;
}
void exgcd(long long a, long long b, long long &x, long long &y, long long &d) {
  if (!b)
    d = a, x = 1, y = 0;
  else
    exgcd(b, a % b, y, x, d), y -= a / b * x;
}
long long inv(long long a, long long p) {
  long long x, y, d;
  exgcd(a, p, x, y, d);
  x = (x % p + p) % p;
  return x;
}
inline int read() {
  char c = getchar();
  int x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
const int maxn = 1e5 + 100;
int n, m, k, t, in[maxn], rk[maxn], w[maxn];
vector<int> g[maxn];
queue<int> q;
int main() {
  cin >> n >> k;
  memset(rk, -1, sizeof(rk));
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
    in[v]++, in[u]++;
  }
  for (int i = 1; i <= n; i++) {
    if (in[i] == 1) {
      q.push(i);
      rk[i] = 0;
    }
  }
  int ok = 0;
  int cnt = 0;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    cnt++;
    for (int v : g[u]) {
      if (in[v] == 1 && rk[v] < rk[u]) continue;
      in[v]--;
      w[v]++;
      if (rk[v] == -1)
        rk[v] = rk[u] + 1;
      else if (rk[v] != rk[u] + 1) {
        puts("No");
        return 0;
      }
      if ((in[v] == 1 && rk[v] != k) || (in[v] == 0 && rk[v] == k)) q.push(v);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (rk[i] == k) ok++;
    if (rk[i] >= 1 && w[i] < 3) ok = 100;
  }
  if (ok == 1 && cnt == n)
    puts("Yes");
  else
    puts("No");
  return 0;
}
