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 INF = 0x3f3f3f3f;
const long double pi = acos(-1);
const int MOD = 1e9 + 7;
const long long int LINF = 0x3f3f3f3f3f3f3f3f;
const int MAXN = 100005;
vector<int> graph[MAXN];
vector<int> g2[MAXN];
int dist[MAXN];
int n, k;
int center;
void dfs(int u, int last) {
  for (int v : graph[u]) {
    if (v == last) continue;
    dist[v] = dist[u] + 1;
    dfs(v, u);
  }
}
bool getCenter(int u, int d, int tgt) {
  if (tgt == u) return true;
  bool found = false;
  for (int v : graph[u]) {
    if (dist[v] > dist[u]) {
      found |= getCenter(v, d, tgt);
    }
  }
  if (found and dist[u] == d) {
    center = u;
    cerr << "found center = " << u << "\n";
  }
  return found;
}
int findCenter() {
  dfs(1, -1);
  int furthest = 1;
  for (int i = 2; i <= n; i++) {
    if (dist[i] > dist[furthest]) {
      furthest = i;
    }
  }
  dist[furthest] = 0;
  dfs(furthest, -1);
  int f2 = 1;
  for (int i = 2; i <= n; i++) {
    if (dist[i] > dist[f2]) {
      f2 = i;
    }
  }
  int d = dist[f2];
  cerr << "d = " << d << "\n";
  if ((d % 2) != 0) {
    puts("No");
    exit(0);
  }
  getCenter(furthest, d / 2, f2);
  return center;
}
bool ok;
void dfs2(int u, int last, int depth) {
  if (graph[u].size() == 1) {
    if (depth != k) {
      puts("No");
      exit(0);
    }
  }
  for (int v : graph[u]) {
    if (v == last) continue;
    dfs2(v, u, depth + 1);
  }
}
int main() {
  ok = true;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  int c = findCenter();
  dist[c] = 0;
  dfs(c, -1);
  if (graph[c].size() < 3) {
    puts("No");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (i == c) continue;
    if (graph[i].size() < 4 and graph[i].size() > 1) {
      puts("No");
      return 0;
    }
  }
  dfs2(c, -1, 0);
  puts("Yes");
  return 0;
}
