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 MAX_N = 1e5 + 7;
int n, k;
std::vector<int> gr[MAX_N];
int dist[MAX_N], pred[MAX_N];
bool good = true;
int yote(const int& cur, const int& dep = 0, const int& par = -1) {
  if (static_cast<int>(gr[cur].size()) <= 1) {
    good = good && dep == k;
    return dep;
  }
  int min = MAX_N, max = -1, vis = 0;
  for (const auto& nei : gr[cur]) {
    if (nei == par) continue;
    int ret = yote(nei, dep + 1, cur);
    min = std::min(min, ret);
    max = std::max(max, ret);
    ++vis;
  }
  good = good && min == max && vis >= 3;
  return dep;
}
int main(void) {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  std::cin >> n >> k;
  for (int i = 1, a, b; i < n; ++i) {
    std::cin >> a >> b;
    --a;
    --b;
    gr[a].push_back(b);
    gr[b].push_back(a);
  }
  memset(dist, 0x3f, sizeof(dist));
  std::queue<int> order;
  order.push(0);
  dist[0] = 0;
  while (order.size()) {
    int cur = order.front();
    order.pop();
    for (const auto& nei : gr[cur]) {
      if (dist[nei] > 1 + dist[cur]) {
        dist[nei] = 1 + dist[cur];
        order.push(nei);
      }
    }
  }
  int mx = std::distance(dist, std::max_element(dist, dist + n));
  memset(dist, 0x3f, sizeof(dist));
  dist[mx] = 0;
  order.push(mx);
  while (order.size()) {
    int cur = order.front();
    order.pop();
    for (const auto& nei : gr[cur]) {
      if (dist[nei] > 1 + dist[cur]) {
        dist[nei] = 1 + dist[cur];
        pred[nei] = cur;
        order.push(nei);
      }
    }
  }
  const int dia = *std::max_element(dist, dist + n);
  if (dia != 2 * k) {
    std::cout << "No\n";
    return 0;
  }
  int center = std::distance(dist, std::max_element(dist, dist + n));
  for (int i = k; i--;) {
    center = pred[center];
  }
  yote(center);
  good = good && n >= 4;
  std::cout << (good ? "Yes" : "No") << "\n";
  return 0;
}
