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 inf = (1 << 30) - 1;
const long long linf = (1ll << 62) - 1;
const int N = 1e5 + 100;
int n, k;
vector<int> g[N];
int dist[N];
bool flag = true;
vector<int> comp;
vector<int> path;
void dfsComp(int v, int p = -1) {
  comp.push_back(v);
  for (int to : g[v]) {
    if (p != to) {
      dfsComp(to, v);
    }
  }
}
void dfs(int v, int p = -1, int h = 0) {
  dist[v] = h;
  for (int to : g[v]) {
    if (p != to) {
      dfs(to, v, h + 1);
    }
  }
}
bool dfsPath(int v, int u, int p = -1) {
  path.push_back(v);
  if (v == u) {
    return true;
  }
  for (int to : g[v]) {
    if (to != p && dfsPath(to, u, v)) {
      return true;
    }
  }
  path.pop_back();
  return false;
}
void gogogo(int v, int k) {
  if (!flag) {
    return;
  }
  comp.clear();
  dfsComp(v);
  dfs(v);
  pair<int, int> best = {-1, -1};
  for (int i : comp) {
    best = max(best, make_pair(dist[i], i));
  }
  v = best.second;
  best = {-1, -1};
  dfs(v);
  for (int i : comp) {
    best = max(best, make_pair(dist[i], i));
  }
  int u = best.second;
  path.clear();
  assert(dfsPath(u, v));
  if (((int)(path).size()) % 2 != 1) {
    flag = false;
    return;
  }
  v = path[((int)(path).size()) / 2];
  if (k == 0 && ((int)(g[v]).size()) == 0) {
    return;
  }
  if (((int)(g[v]).size()) < 3) {
    flag = false;
    return;
  }
  for (int to : g[v]) {
    g[to].erase(find((g[to]).begin(), (g[to]).end(), v));
    gogogo(to, k - 1);
  }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v), u--, v--;
    g[u].push_back(v), g[v].push_back(u);
  }
  gogogo(0, k);
  if (flag) {
    puts("Yes");
  } else {
    puts("No");
  }
  return 0;
}
