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 double PI = acos(-1.0);
const int INF = 1000 * 1000 * 1000 + 7;
const long long LINF = (long long)INF * INF;
const int MAX = 100 * 1000 + 7;
vector<int> g[MAX];
int d[4][MAX];
void dfs(int v, int p, int idx) {
  for (int i = (0); i < ((int)((g[v]).size())); ++i) {
    int to = g[v][i];
    if (to != p) {
      d[idx][to] = d[idx][v] + 1;
      dfs(to, v, idx);
    }
  }
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = (0); i < (n - 1); ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0, -1, 0);
  int u = 0;
  for (int i = (1); i < (n); ++i)
    if (d[0][i] > d[0][u]) u = i;
  dfs(u, -1, 1);
  int v = 0;
  for (int i = (1); i < (n); ++i)
    if (d[1][i] > d[1][v]) v = i;
  dfs(v, -1, 2);
  int w = -1;
  for (int i = (0); i < (n); ++i)
    if (d[1][i] == d[2][i] && (w == -1 || d[1][i] < d[1][w])) {
      w = i;
    }
  if (w == -1 || (int)((g[w]).size()) < 3) {
    printf("No");
    return 0;
  }
  dfs(w, -1, 3);
  for (int i = (0); i < (n); ++i) {
    if ((int)((g[i]).size()) == 1) {
      if (d[3][i] != k) {
        printf("No");
        return 0;
      }
    } else if (i != w && (int)((g[i]).size()) < 4) {
      printf("No");
      return 0;
    }
  }
  printf("Yes");
  return 0;
}
