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 N = 100050;
int dep[N], par[N];
vector<int> E[N];
void DFS(int u, int p) {
  dep[u] = dep[p] + 1;
  par[u] = p;
  for (int v : E[u])
    if (v != p) DFS(v, u);
}
bool ok = 1;
void Check(int u, int p, int d, int k) {
  int ch = 0;
  for (int v : E[u])
    if (v != p) {
      ch++;
      Check(v, u, d + 1, k);
    }
  if (ch == 0) {
    if (d != k + 1) ok = 0;
  } else if (ch < 3)
    ok = 0;
}
int main() {
  int n, k, i, u, v;
  scanf("%i %i", &n, &k);
  for (i = 1; i < n; i++)
    scanf("%i %i", &u, &v), E[u].push_back(v), E[v].push_back(u);
  DFS(1, 0);
  int cen = 0;
  for (i = 1; i <= n; i++)
    if (dep[cen] < dep[i]) cen = i;
  DFS(cen, 0);
  int den = 0;
  for (i = 1; i <= n; i++)
    if (dep[den] < dep[i]) den = i;
  int goal = (dep[den] + 1) / 2;
  int ctr = den;
  while (dep[ctr] > goal) ctr = par[ctr];
  Check(ctr, 0, 1, k);
  if (ok)
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}
