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;
void qmax(int &x, int y) {
  if (x < y) x = y;
}
void qmin(int &x, int y) {
  if (x > y) x = y;
}
inline int read() {
  char s;
  int k = 0, base = 1;
  while ((s = getchar()) != '-' && s != EOF && !(isdigit(s)))
    ;
  if (s == EOF) exit(0);
  if (s == '-') base = -1, s = getchar();
  while (isdigit(s)) k = k * 10 + (s ^ '0'), s = getchar();
  return k * base;
}
inline void write(long long x) {
  static char cnt, num[20];
  cnt = 0;
  if (!x) {
    putchar('0');
    return;
  }
  for (; x; x /= 10) num[++cnt] = x % 10;
  for (; cnt; putchar(num[cnt--] + 48))
    ;
}
const int maxn = 1e5 + 100;
int po[maxn], to[maxn << 1], ne[maxn << 1], id;
int d[maxn], dis[maxn], c[maxn], fa[maxn];
void add(int x, int y) {
  id++;
  to[id] = y, ne[id] = po[x], po[x] = id;
}
int n, k;
queue<int> q;
int main() {
  n = read(), k = read();
  for (int i = 1, x, y; i < n; i++) {
    x = read(), y = read();
    d[x]++, d[y]++;
    add(x, y);
    add(y, x);
  }
  memset(dis, -1, sizeof(dis));
  int u, v, cnt = 0;
  for (int i = 1; i <= n; i++)
    if (d[i] == 1) {
      q.push(i), dis[i] = 0, cnt++;
    }
  for (int i = 1; i <= n; i++)
    if (d[i] == 1) {
      for (int j = po[i]; j; j = ne[j])
        if (d[to[j]] != 1) {
          fa[i] = to[j];
          break;
        }
    }
  while (!q.empty()) {
    u = q.front();
    q.pop();
    v = fa[u];
    c[v]++;
    d[v]--;
    if (dis[v] == -1 || dis[v] == dis[u] + 1)
      dis[v] = dis[u] + 1;
    else {
      printf("No");
      return 0;
    };
    if (cnt == n - 1 && q.empty()) {
      if (d[v] == 0 && c[v] >= 3 && dis[v] == k) {
        printf("Yes");
        return 0;
      } else {
        printf("No");
        return 0;
      }
    }
    if (d[v] == 1 && cnt != n - 1) {
      if (c[v] < 3) {
        printf("No");
        return 0;
      }
      for (int i = po[v]; i; i = ne[i])
        if (dis[to[i]] == -1) {
          fa[v] = to[i];
          break;
        }
      q.push(v);
      cnt++;
      dis[v] = dis[u] + 1;
    }
  }
  {
    printf("No");
    return 0;
  }
  return 0;
}
