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 inf = 0x7fffffff;
const double eps = 1e-10;
const double pi = acos(-1.0);
inline int read() {
  int x = 0, f = 1;
  char ch;
  ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch & 15);
    ch = getchar();
  }
  if (f)
    return x;
  else
    return -x;
}
const int N = 1e5 + 10;
int n, K;
int head[N], p, du[N], lv[N];
struct ss {
  int v, last;
} G[N << 1];
void add(int a, int b) {
  du[a]++;
  du[b]++;
  G[++p] = (ss){b, head[a]};
  head[a] = p;
  G[++p] = (ss){a, head[b]};
  head[b] = p;
}
queue<int> Q;
int cnt, rt;
bool dfs(int a, int fa, int k) {
  int size = 0, f = 1;
  for (int i = head[a]; i; i = G[i].last)
    if (G[i].v != fa) {
      ++size;
      if (!dfs(G[i].v, a, k - 1)) f = 0;
    }
  if (k == 0)
    return !size;
  else
    return f && size >= 3;
}
int main() {
  n = read();
  K = read();
  for (int i = 1, a, b; i < n; i++) {
    a = read();
    b = read();
    add(a, b);
  }
  for (int i = 1; i <= n; i++)
    if (du[i] == 1) {
      Q.push(i);
      lv[i] = 0;
    }
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    if ((++cnt) == n) {
      rt = u;
      break;
    }
    for (int i = head[u]; i; i = G[i].last) {
      du[G[i].v]--;
      if (du[G[i].v] == 1) Q.push(G[i].v);
    }
  }
  printf("%s\n", dfs(rt, 0, K) ? "Yes" : "No");
  return 0;
}
