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 = 1000000009;
const long long LINF = 1000000000000000009LL;
const int N = 100005;
int n, k, a, b;
struct Vertex : vector<int> {
  int dist, odw;
};
Vertex g[N];
int dist = 0;
void dfs(int v, int u) {
  g[v].odw = u;
  g[v].dist = dist++;
  for (auto w : g[v]) {
    if (w != u) {
      dfs(w, v);
    }
  }
  dist--;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i <= n - 1 - 1; ++i) {
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(1, -1);
  int poc = 1, kon = 1, sr = 1;
  for (int i = 1; i <= n; ++i)
    if (g[i].dist > g[poc].dist) {
      poc = i;
    }
  dfs(poc, -1);
  for (int i = 1; i <= n; ++i)
    if (g[i].dist > g[kon].dist) {
      kon = i;
    }
  bool res = false;
  if (g[kon].dist % 2 == 0 && g[kon].dist / 2 == k) {
    res = true;
    sr = kon;
    int hdg = g[kon].dist / 2;
    for (int i = 0; i <= hdg - 1; ++i) {
      sr = g[sr].odw;
    }
    dfs(sr, -1);
    if ((int)(g[sr]).size() < 3) {
      res = false;
    }
    for (int i = 1; i <= n; ++i)
      if ((int)(g[i]).size() == 1 && g[i].dist != hdg ||
          ((int)(g[i]).size() == 2 || (int)(g[i]).size() == 3) && i != sr) {
        res = false;
        break;
      }
  }
  cout << (res ? "Yes\n" : "No\n");
  return 0;
}
