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 head[N], cnt = 0;
struct Edge {
  int v, next;
} edge[N << 1];
queue<int> q;
int d[N], dis[N];
void add(int u, int v) {
  edge[++cnt] = (Edge){v, head[u]};
  head[u] = cnt;
}
int main() {
  int n, k;
  memset(dis, 120, sizeof(dis));
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(u, v), add(v, u);
    d[u]++, d[v]++;
  }
  for (int i = 1; i <= n; i++)
    if (d[i] == 1) dis[i] = 0, q.push(i);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = head[x]; i; i = edge[i].next)
      if (dis[edge[i].v] > dis[x] + 1) {
        dis[edge[i].v] = dis[x] + 1;
        q.push(edge[i].v);
      }
  }
  for (int i = 1; i <= n; i++) {
    int cnt1 = 0, cnt2 = 0;
    for (int j = head[i]; j; j = edge[j].next)
      if (dis[edge[j].v] == dis[i]) {
        puts("NO");
        return 0;
      } else if (dis[edge[j].v] == dis[i] - 1)
        cnt1++;
      else
        cnt2++;
    if ((dis[i] && cnt1 < 3) || (dis[i] != k && cnt2 != 1)) {
      puts("NO");
      return 0;
    }
  }
  puts("YES");
  return 0;
}
