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;
using Long = long long;
vector<vector<int>> G(100010);
int D[100010];
int dist[100010];
int P[100010];
int ep1 = -1, ep1lvl = -1, ep2 = -1, ep2lvl = -1;
void dfs(int u, int p, int lvl) {
  if (lvl > ep1lvl) {
    ep1lvl = lvl;
    ep1 = u;
  }
  for (auto v : G[u]) {
    if (v != p) dfs(v, u, lvl + 1);
  }
}
void dfs1(int u, int p, int lvl) {
  P[u] = p;
  if (lvl > ep2lvl) {
    ep2lvl = lvl;
    ep2 = u;
  }
  for (auto v : G[u]) {
    if (v != p) dfs1(v, u, lvl + 1);
  }
}
int md;
vector<int> path;
void getMid(int u) {
  while (P[u] != -1) {
    path.push_back(u);
    u = P[u];
  }
  path.push_back(u);
  reverse(path.begin(), path.end());
}
bool ok = true;
int n, k;
int main() {
  cin >> n >> k;
  memset(D, 0, sizeof(D));
  for (int i = 0; i < 100000; i++) P[i] = -1, dist[i] = 1 << 30;
  int first = 0;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    if (!i) first = u;
    G[u].push_back(v);
    G[v].push_back(u);
    D[u]++, D[v]++;
  }
  dfs(first, -1, 1);
  dfs1(ep1, -1, 1);
  if (ep2lvl <= k) {
    puts("No");
    return 0;
  }
  getMid(ep2);
  md = path[k];
  queue<int> Q;
  Q.push(md);
  dist[md] = 0;
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    if (u == md) {
      if (D[u] < 3) ok = false;
    } else {
      if (dist[u] != k && (D[u] - 1) < 3) ok = false;
      if (dist[u] == k && D[u] > 1) ok = false;
    }
    for (int v : G[u]) {
      if (dist[v] == 1 << 30) {
        dist[v] = dist[u] + 1;
        Q.push(v);
      }
    }
  }
  (ok) ? puts("Yes") : puts("No");
  return 0;
}
