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 N, K;
int ui, vi;
vector<pair<int, int> > G[100001];
int maxdis[100001];
pair<int, int> dis[100001];
void dfs(int node, int pa) {
  dis[node] = make_pair(0, 0);
  for (int i = 0; i < G[node].size(); i++) {
    int child = G[node][i].first;
    if (child == pa) {
      continue;
    }
    dfs(child, node);
    if (dis[child].first + G[node][i].second >= dis[node].first) {
      dis[node] =
          make_pair(dis[child].first + G[node][i].second, dis[node].first);
    } else if (dis[child].first + G[node][i].second > dis[node].second) {
      dis[node] =
          make_pair(dis[node].first, dis[child].first + G[node][i].second);
    }
  }
}
void dfs1(int node, int pa) {
  for (int i = 0; i < G[node].size(); i++) {
    int child = G[node][i].first;
    if (child == pa) {
      continue;
    }
    if (dis[node].first != dis[node].second &&
        dis[child].first + G[node][i].second == dis[node].first) {
      maxdis[child] = max(maxdis[node] + G[node][i].second,
                          dis[node].second + G[node][i].second);
    } else {
      maxdis[child] = max(maxdis[node] + G[node][i].second,
                          dis[node].first + G[node][i].second);
    }
    dfs1(child, node);
  }
}
bool visited[100001];
bool check(int node, int depth) {
  visited[node] = true;
  bool ismhh = true;
  if (depth == K) {
    if (G[node].size() != 1) {
      ismhh = false;
    }
  } else {
    if ((depth == 0 && G[node].size() < 3) ||
        (depth != 0 && G[node].size() < 4)) {
      ismhh = false;
    }
  }
  for (int i = 0; i < G[node].size(); i++) {
    if (!visited[G[node][i].first]) {
      if (!check(G[node][i].first, depth + 1)) {
        ismhh = false;
        break;
      }
    }
  }
  return ismhh;
}
int maxdist[100001];
int main() {
  scanf("%d %d", &N, &K);
  for (int i = 0; i < N - 1; i++) {
    scanf("%d %d", &ui, &vi);
    G[ui].push_back({vi, 1});
    G[vi].push_back({ui, 1});
  }
  dfs(1, -1);
  dfs1(1, -1);
  for (int i = 1; i <= N; i++) {
    maxdist[i] = max(maxdis[i], dis[i].first);
  }
  int best = 1;
  for (int i = 2; i <= N; i++) {
    if (maxdist[i] < maxdist[best]) {
      best = i;
    }
  }
  if (check(best, 0)) {
    printf("Yes\n");
  } else {
    printf("No\n");
  }
}
