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 maxn = 100001;
vector<int> adj[maxn];
bool vis[maxn];
int dia[maxn];
int dp[maxn];
int n, k;
int find_dia();
void cal(int u);
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  int m = find_dia();
  if (m % 2 == 0) {
    printf("No");
    return 0;
  }
  int g = dia[m / 2 + 1];
  cal(g);
  printf("%s", dp[g] == k ? "Yes" : "No");
  return 0;
}
int dis[maxn], prv[maxn];
void dfs(int u) {
  dis[u] = 0;
  prv[u] = 0;
  vis[u] = 1;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (vis[v]) continue;
    dfs(v);
    if (dis[v] + 1 > dis[u]) {
      dis[u] = dis[v] + 1;
      prv[u] = v;
    }
  }
  vis[u] = 0;
}
int find_dia() {
  dfs(1);
  int s = 1;
  while (prv[s]) s = prv[s];
  dfs(s);
  int ans = dis[s];
  for (int i = 1; i <= ans + 1; i++) {
    dia[i] = s;
    s = prv[s];
  }
  return ans + 1;
}
void cal(int u) {
  vis[u] = 1;
  if (adj[u].size() == 1) return;
  int ch = 0;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (vis[v]) continue;
    ch++;
    cal(v);
    if (dp[v] == -1) {
      dp[u] = -1;
      break;
    }
    if (ch == 1) {
      dp[u] = dp[v] + 1;
    } else if (dp[v] + 1 != dp[u]) {
      dp[u] = -1;
      break;
    }
  }
  if (ch < 3) dp[u] = -1;
}
