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 t, n, m, k, s, minn, maxx, ans, sum, a[100005], b[100005], cnt[100005], pos,
    pos2 = -1, dp1[100005], dp2[100005], dp3[100005], vis[100005], vis2[100005],
    vis3[100005], pos3 = -1;
vector<int> v[100005];
void dfs(int x) {
  vis[x] = 1;
  for (int ii = 0; ii < cnt[x]; ii++) {
    if (vis[v[x][ii]] == 0) {
      dp1[v[x][ii]] = dp1[x] + 1;
      dfs(v[x][ii]);
    }
  }
}
void dfs2(int x) {
  vis2[x] = 1;
  for (int ii = 0; ii < cnt[x]; ii++) {
    if (vis2[v[x][ii]] == 0) {
      dp2[v[x][ii]] = dp2[x] + 1;
      dfs2(v[x][ii]);
    }
  }
}
void dfs3(int x) {
  vis3[x] = 1;
  for (int ii = 0; ii < cnt[x]; ii++) {
    if (vis3[v[x][ii]] == 0) {
      dp3[v[x][ii]] = dp3[x] + 1;
      dfs3(v[x][ii]);
    }
  }
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  for (int i = 0; i < 100002; i++) {
    cnt[i] = 0;
    vis[i] = 0;
    vis2[i] = 0;
    vis3[i] = 0;
  }
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    cin >> a[i] >> b[i];
    cnt[a[i]]++;
    cnt[b[i]]++;
    v[a[i]].push_back(b[i]);
    v[b[i]].push_back(a[i]);
  }
  for (int i = 1; i <= n; i++) {
    if (cnt[i] == 1) {
      pos = i;
      break;
    }
  }
  dp1[pos] = 0;
  dfs(pos);
  for (int i = 1; i <= n; i++) {
    if (dp1[i] > 2 * k) {
      cout << "NO" << '\n';
      return 0;
    }
    if (dp1[i] == 2 * k) {
      pos2 = i;
    }
  }
  if (pos2 == -1) {
    cout << "NO" << '\n';
    return 0;
  }
  dp2[pos2] = 0;
  dfs2(pos2);
  for (int i = 1; i <= n; i++) {
    if (dp1[i] == k && dp2[i] == k) {
      pos3 = i;
      break;
    }
  }
  if (pos3 == -1) {
    cout << "NO" << '\n';
    return 0;
  } else {
    dp3[pos3] = 0;
    dfs3(pos3);
    for (int i = 1; i <= n; i++) {
      if (dp3[i] < k && dp3[i] > 0) {
        if (cnt[i] < 4) {
          cout << "NO" << '\n';
          return 0;
        }
      } else if (dp3[i] == k) {
        if (cnt[i] > 1) {
          cout << "NO" << '\n';
          return 0;
        }
      } else {
        if (cnt[i] < 3) {
          cout << "NO" << '\n';
          return 0;
        }
      }
    }
  }
  cout << "YES" << '\n';
}
