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 N = 1e5 + 5;
int n, K;
struct edge {
  int to, next;
} e[N << 1];
int head[N], cnt, num;
int vis[N];
int in[N];
inline void add(int x, int y) {
  e[++cnt].to = y;
  e[cnt].next = head[x];
  head[x] = cnt;
}
inline void solve() {
  queue<int> q[2];
  for (int i = 1; i <= n; i++)
    if (in[i] == 1) q[0].push(i), vis[i] = 0;
  for (int t = 0;; t ^= 1) {
    while (!q[t].empty()) {
      int x = q[t].front();
      q[t].pop();
      if (in[x] == 0) continue;
      int u;
      for (int i = head[x]; i; i = e[i].next) {
        if (in[e[i].to] == 0) continue;
        u = e[i].to;
        break;
      }
      if (in[u] < 3) {
        printf("No\n");
        exit(0);
      }
      int tot = 0;
      for (int i = head[u]; i; i = e[i].next) {
        int v = e[i].to;
        if (in[v] == 1 && vis[v] == t) in[v]--, in[u]--, tot++;
      }
      if (tot < 3) {
        printf("No\n");
        exit(0);
      }
      q[t ^ 1].push(u);
      vis[u] = t ^ 1;
    }
    if (q[t ^ 1].empty()) break;
    num++;
  }
}
int main() {
  scanf("%d%d", &n, &K);
  memset(vis, -1, sizeof(vis));
  for (int i = 1, x, y; i < n; i++) {
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
    in[x]++;
    in[y]++;
  }
  if (K >= n) {
    printf("No\n");
    return 0;
  }
  solve();
  if (num == K)
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}
