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 md;
int mdv;
bool used[200010];
vector<int> e[200010];
int dd[200010];
int dd2[200010];
void dfs(int k, int d) {
  used[k] = true;
  dd[k] = d;
  if (md < d) {
    md = d;
    mdv = k;
  }
  for (int i = 0; i < (int)e[k].size(); i++) {
    if (!used[e[k][i]]) {
      dfs(e[k][i], d + 1);
    }
  }
  return;
}
int main(void) {
  int n, k;
  int x, y;
  int c;
  scanf("%d %d", &n, &k);
  for (int i = 0; i < (int)n - 1; i++) {
    scanf("%d %d", &x, &y);
    e[x - 1].push_back(y - 1);
    e[y - 1].push_back(x - 1);
  }
  md = -1;
  for (int i = 0; i < (int)n; i++) used[i] = false;
  dfs(0, 0);
  md = -1;
  for (int i = 0; i < (int)n; i++) used[i] = false;
  dfs(mdv, 0);
  if (md != (2 * k)) {
    printf("No\n");
    return 0;
  }
  for (int i = 0; i < (int)n; i++) dd2[i] = dd[i];
  md = -1;
  for (int i = 0; i < (int)n; i++) used[i] = false;
  dfs(mdv, 0);
  for (int i = 0; i < (int)n; i++) {
    if (dd[i] == k && dd2[i] == k) {
      c = i;
      break;
    }
  }
  md = -1;
  for (int i = 0; i < (int)n; i++) used[i] = false;
  dfs(c, 0);
  for (int i = 0; i < (int)n; i++) {
    if (i == c) {
      if (e[i].size() < 3) {
        printf("No\n");
        return 0;
      }
    } else {
      if (e[i].size() == 1) {
        if (dd[i] != k) {
          printf("No\n");
          return 0;
        }
      } else if (e[i].size() < 4) {
        printf("No\n");
        return 0;
      }
    }
  }
  printf("Yes\n");
  return 0;
}
