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 = 100005;
bool vis[MAXN];
bool inQueue[MAXN];
int layer[MAXN];
int sons[MAXN];
bool yes = true;
vector<int> graph[MAXN];
int main() {
  int ans = 0, n, k;
  scanf("%d%d", &n, &k);
  queue<int> que;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) {
    if (graph[i].size() == 1) {
      que.push(i);
    }
  }
  int nums = 0;
  while (!que.empty()) {
    int u = que.front();
    vis[u] = true;
    nums++;
    que.pop();
    if (sons[u] < 3 && layer[u] != 0) {
      yes = false;
      break;
    }
    int cnt = 0;
    int v = 0;
    for (int i = 0; i < graph[u].size(); i++) {
      if (!vis[graph[u][i]]) {
        cnt++;
        v = graph[u][i];
      }
    }
    if (cnt != 1 && nums != n) {
      yes = false;
      break;
    }
    if (v == 0) {
      break;
    }
    if (layer[v] != 0 && layer[v] != layer[u] + 1) {
      yes = false;
      break;
    }
    layer[v] = layer[u] + 1;
    ans = max(ans, layer[v]);
    sons[v]++;
    if (!inQueue[v]) {
      que.push(v);
      inQueue[v] = true;
    }
  }
  if (!yes || ans != k) {
    cout << "No" << endl;
  } else {
    cout << "Yes" << endl;
  }
  return 0;
}
