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;
struct Edge {
  int v;
  Edge* next;
};
const int N = 100010;
int n, k;
bool vis[N];
int dis[N], prt[N], deg[N];
Edge* g[N];
void adde(int u, int v);
void dfs(int u);
void pred(int u, int r);
int main() {
  scanf("%d%d", &n, &k);
  for (int rep = 1; rep < n; ++rep) {
    int u, v;
    scanf("%d%d", &u, &v);
    adde(u, v);
    adde(v, u);
  }
  dfs(1);
  int u = max_element(dis + 1, dis + n + 1) - dis;
  memset(vis, 0, sizeof(vis));
  dis[u] = 0;
  prt[u] = 0;
  dfs(u);
  int m = max_element(dis + 1, dis + n + 1) - dis;
  if (dis[m] != k * 2) {
    puts("No");
    return 0;
  }
  int cnt = k;
  while (cnt--) m = prt[m];
  memset(vis, 0, sizeof(vis));
  pred(m, k);
  puts("YES");
  return 0;
}
void pred(int u, int r) {
  vis[u] = true;
  if (r == 0) {
    if (deg[u]) {
      puts("NO");
      exit(0);
    }
    return;
  }
  if (deg[u] < 3) {
    puts("NO");
    exit(0);
  }
  for (Edge* p = g[u]; p; p = p->next)
    if (!vis[p->v]) {
      --deg[p->v];
      pred(p->v, r - 1);
    }
}
void dfs(int u) {
  vis[u] = true;
  for (Edge* p = g[u]; p; p = p->next)
    if (!vis[p->v]) {
      prt[p->v] = u;
      dis[p->v] = dis[u] + 1;
      dfs(p->v);
    }
}
void adde(int u, int v) {
  static Edge pool[N * 2];
  static Edge* p = pool;
  p->v = v;
  p->next = g[u];
  ++deg[u];
  g[u] = p;
  ++p;
}
