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>
const int maxn = 1e5 + 7;
using namespace std;
int n, k, root, ans, cnt;
int ls[maxn], son[maxn], r[maxn], dis[maxn], f[maxn];
struct edge {
  int y, next;
} g[maxn * 2];
void add(int x, int y) {
  g[++cnt] = (edge){y, ls[x]};
  ls[x] = cnt;
}
void findroot(int x, int fa) {
  f[x] = fa;
  for (int i = ls[x]; i > 0; i = g[i].next) {
    int y = g[i].y;
    if (y == fa) continue;
    dis[y] = dis[x] + 1;
    findroot(y, x);
  }
}
void dfs(int x, int fa, int dep) {
  for (int i = ls[x]; i > 0; i = g[i].next) {
    int y = g[i].y;
    if (y == fa) continue;
    son[x]++;
    dfs(y, x, dep + 1);
  }
  if (dep > k) ans = 0;
  if ((dep == k) && (son[x])) ans = 0;
  if ((dep < k) && (son[x] <= 2)) ans = 0;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
    r[x]++, r[y]++;
  }
  for (int i = 1; i <= n; i++) {
    if (r[i] == 1) {
      root = i;
      break;
    }
  }
  findroot(root, 0);
  for (int i = 1; i <= n; i++) {
    if (dis[i] > dis[root]) root = i;
  }
  ans = 1;
  if (dis[root] != k * 2)
    ans = 0;
  else {
    for (int i = 1; i <= k; i++) root = f[root];
    dfs(root, 0, 0);
  }
  if (ans)
    printf("Yes");
  else
    printf("No");
}
