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>
#pragma warning(disable : 4996)
using namespace std;
const int maxn = 200005;
int head[maxn], nex[maxn], ver[maxn];
int n, k;
int son[maxn];
int fa[maxn];
int dis[maxn];
int cnt;
void add(int x, int y) {
  ver[cnt] = y;
  nex[cnt] = head[x];
  head[x] = cnt++;
}
int dfs1(int x, int f) {
  fa[x] = f;
  int ans = x;
  for (int i = head[x]; i != -1; i = nex[i]) {
    int v = ver[i];
    if (v == f) {
      continue;
    }
    dis[v] = dis[x] + 1;
    int p = dfs1(v, x);
    if (dis[p] > dis[ans]) {
      ans = p;
    }
  }
  return ans;
}
void dfs2(int rt, int f) {
  for (int i = head[rt]; i != -1; i = nex[i]) {
    int v = ver[i];
    if (v == f) {
      continue;
    }
    dis[v] = dis[rt] + 1;
    son[rt]++;
    dfs2(v, rt);
  }
}
int main() {
  scanf("%d%d", &n, &k);
  memset(head, -1, sizeof(head));
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    add(a, b);
    add(b, a);
  }
  memset(dis, 0, sizeof(dis));
  fa[0] = 0;
  int x = dfs1(1, 0);
  memset(dis, 0, sizeof(dis));
  int rt = dfs1(x, 0);
  if (dis[rt] != 2 * k) {
    printf("No\n");
  } else {
    for (int i = 1; i <= k; i++) {
      rt = fa[rt];
    }
    memset(dis, 0, sizeof(dis));
    memset(son, 0, sizeof(son));
    dfs2(rt, 0);
    bool flag = true;
    for (int i = 1; i <= n; i++) {
      if (dis[i] != k && son[i] == 0) {
        flag = false;
        break;
      } else if (son[i] >= 1 && son[i] < 3) {
        flag = false;
        break;
      }
    }
    if (flag) {
      printf("Yes\n");
    } else {
      printf("No\n");
    }
  }
  return 0;
}
