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;
inline char nc() {
  static char buf[100000], *p1, *p2;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
template <class T>
inline void rd(T& x) {
  x = 0;
  char ch = nc();
  for (; !isdigit(ch); ch = nc())
    ;
  for (; isdigit(ch); ch = nc()) x = x * 10 - 48 + ch;
}
const int maxn = 1e5 + 10;
const int inf = 0x3f3f3f3f;
struct edge {
  int v, next;
  edge(int v = 0, int next = 0) : v(v), next(next) {}
} e[maxn * 2];
int hd[maxn], ek = 1;
inline void adde(int u, int v) {
  e[ek] = edge(v, hd[u]);
  hd[u] = ek++;
}
int fa[maxn], dep[maxn], n, cent;
void dfs(int u) {
  for (int i = hd[u]; i; i = e[i].next) {
    int v = e[i].v;
    if (v == fa[u]) continue;
    dep[v] = dep[u] + 1;
    fa[v] = u;
    dfs(v);
  }
}
void getcent() {
  dfs(1);
  int mx = -1, ms = 0;
  for (int i = 1; i <= n; i++) {
    if (dep[i] > mx) {
      mx = dep[i];
      ms = i;
    }
  }
  memset(dep, 0, sizeof(dep));
  memset(fa, 0, sizeof(fa));
  dfs(ms);
  mx = -1, ms = 0;
  for (int i = 1; i <= n; i++) {
    if (dep[i] > mx) {
      mx = dep[i];
      ms = i;
    }
  }
  for (cent = ms; dep[cent] != dep[ms] / 2; cent = fa[cent])
    ;
}
int deg[maxn];
bool check(int u, int fa, int k) {
  if (k == 0 && deg[u] == 0) return true;
  if (deg[u] < 3) return false;
  for (int i = hd[u]; i; i = e[i].next) {
    int v = e[i].v;
    if (v == fa) continue;
    deg[v]--;
    if (!check(v, u, k - 1)) return 0;
  }
  return true;
}
int main() {
  int k;
  rd(n);
  rd(k);
  for (int i = 1; i < n; i++) {
    int u, v;
    rd(u);
    rd(v);
    adde(u, v);
    adde(v, u);
    deg[u]++;
    deg[v]++;
  }
  getcent();
  printf("%s\n", check(cent, cent, k) ? "Yes" : "No");
}
