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 = 1e5 + 10;
int n, m, sum, d[N], e[N], f[N], head[N];
inline int read() {
  int x = 0, p = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') p = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * p;
}
struct ROAD {
  int des, last;
} road[N << 1];
void add(int u, int v) {
  road[++sum].des = v;
  road[sum].last = head[u];
  head[u] = sum;
}
queue<int> zy;
bool topo() {
  memset(e, -1, sizeof e);
  for (int i = 1; i <= n; ++i)
    if (d[i] == 1) {
      e[i] = m;
      zy.push(i);
    }
  while (!zy.empty()) {
    int u = zy.front();
    zy.pop();
    if (e[u] < m && e[u] > 0 && f[u] <= 3) return false;
    if (e[u] == 0 && f[u] < 3) return false;
    for (int i = head[u]; i; i = road[i].last) {
      int v = road[i].des;
      if (--d[v] == 1) zy.push(v);
      if (d[v] != 0)
        if (e[v] == -1)
          e[v] = e[u] - 1;
        else if (e[v] != e[u] - 1)
          return false;
    }
  }
  int ret = 0;
  for (int i = 1; i <= n; ++i) {
    if (e[i] == -1) return false;
    ret += (e[i] == 0);
  }
  return ret == 1;
}
int main() {
  n = read();
  m = read();
  if (m > 447) return puts("No"), 0;
  for (int i = 1; i < n; ++i) {
    int u = read(), v = read();
    ++d[u];
    ++d[v];
    ++f[u];
    ++f[v];
    add(u, v);
    add(v, u);
  }
  if (topo())
    puts("Yes");
  else
    puts("No");
  return 0;
}
