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;
int const N = 1e5 + 5, M = 8, OO = 0x3f3f3f3f;
int n, k, u, v, ans, center;
vector<int> adj[N];
int degCenter[N], deg[N];
bool res = 1;
bool vis[N];
int getCenter() {
  int remain = n, u, level;
  queue<int> q;
  for (int i = 1; i <= n; i++) {
    if (deg[i] == 1) {
      q.push(i);
      deg[i] = 0;
    }
  }
  while (remain > 1) {
    level = q.size();
    if (level) ans++;
    while (level--) {
      u = q.front();
      q.pop();
      remain--;
      for (int v : adj[u]) {
        deg[v]--;
        if (deg[v] == 1) q.push(v);
      }
    }
  }
  return q.front();
}
void BFS(int src) {
  int u, temp = 0, level = 0, cntlevelo = 1, cntleveln = 0, cnt = 0;
  queue<int> q;
  vis[src] = 1;
  q.push(src);
  while (!q.empty()) {
    u = q.front();
    q.pop();
    if (cntlevelo > cnt) {
      cntleveln += degCenter[u] - 1;
      if (u == src) cntleveln++;
    } else {
      level++;
      cntlevelo = cntleveln;
      cntleveln = degCenter[u] - 1;
      cnt = 0;
    }
    cnt++;
    if (level == 0 && degCenter[u] < 3) {
      res = 0;
    } else if (level == k && degCenter[u] != 1) {
      res = 0;
    } else if (level > 0 && level < k && degCenter[u] < 4) {
      res = 0;
    }
    for (int v : adj[u]) {
      if (!vis[v]) {
        vis[v] = 1;
        q.push(v);
      }
    }
  }
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &u, &v);
    degCenter[u]++, deg[u]++;
    degCenter[v]++, deg[v]++;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  center = getCenter();
  if (ans == k) {
    BFS(center);
    if (res)
      puts("Yes");
    else
      puts("No");
  } else
    puts("No");
  return 0;
}
