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 N, K;
const int MAXN = 1e5 + 10;
class Edges {
 public:
  int next, to;
};
Edges edge[MAXN << 1];
int first[MAXN];
int deg[MAXN];
int te = 1;
void add(int s, int t) {
  ++te;
  edge[te].to = t;
  edge[te].next = first[s];
  first[s] = te;
  deg[s]++;
}
int len[MAXN];
bool flag[MAXN];
void dfs(int x) {
  flag[x] = 1;
  for (int e = first[x]; e; e = edge[e].next) {
    int to = edge[e].to;
    if (!flag[to]) {
      len[to] = len[x] + 1;
      dfs(to);
    }
  }
}
bool check(int pt, int l) {
  if (l == 0) return 1;
  if (deg[pt] < 3) return 0;
  flag[pt] = 1;
  bool ret = 1;
  for (int e = first[pt]; e; e = edge[e].next) {
    int to = edge[e].to;
    if (!flag[to]) {
      deg[to]--;
      ret &= check(to, l - 1);
    }
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> N >> K;
  for (int i = 1; i < N; ++i) {
    int u, v;
    cin >> u >> v;
    add(u, v);
    add(v, u);
  }
  memset(flag, 0, sizeof flag);
  len[1] = 0;
  dfs(1);
  int wh = 1;
  for (int i = 1; i <= N; ++i)
    if (len[i] > len[wh]) wh = i;
  memset(flag, 0, sizeof flag);
  len[wh] = 0;
  dfs(wh);
  int st = wh;
  for (int i = 1; i <= N; ++i)
    if (len[i] > len[wh]) wh = i;
  if (len[wh] % 2 != 0 || len[wh] / 2 != K) {
    cout << "No" << endl;
    return 0;
  }
  int MID = wh;
  while (len[MID] != len[wh] / 2) {
    for (int e = first[MID]; e; e = edge[e].next) {
      int to = edge[e].to;
      if (len[to] < len[MID]) {
        MID = to;
        break;
      }
    }
  }
  memset(flag, 0, sizeof flag);
  if (check(MID, K))
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
  return 0;
}
