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;
int n, k;
vector<int> g[100010];
const int inf = 1e9;
int d[100010], p[100010];
queue<int> Q;
void bfs(int x) {
  for (int i = 1; i <= n; i++) d[i] = inf;
  d[x] = 0;
  while (!Q.empty()) Q.pop();
  Q.push(x);
  while (!Q.empty()) {
    int y = Q.front();
    Q.pop();
    for (int z : g[y]) {
      if (d[z] < inf) continue;
      d[z] = d[y] + 1;
      p[z] = y;
      Q.push(z);
    }
  }
}
bool dfs(int x, int pa, int c) {
  int d = g[x].size();
  if (pa) d--;
  if (c) {
    if (d < 3) return 0;
  } else {
    if (d) return 0;
  }
  for (int y : g[x]) {
    if (y == pa) continue;
    if (!dfs(y, x, c - 1)) return 0;
  }
  return 1;
}
int main() {
  cin >> n >> k;
  for (int i = 1, x, y; i < n; i++) {
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  bfs(1);
  int x = max_element(d + 1, d + n + 1) - d;
  bfs(x);
  int y = max_element(d + 1, d + n + 1) - d;
  int l = d[y];
  while (2 * d[y] > l) y = p[y];
  puts(dfs(y, 0, k) ? "Yes" : "No");
}
