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.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

/**
 * @author Don Li
 */
public class Multihedgehog {
    
    int[] dep, par;
    int[][] G;
    
    void solve() {
        int n = in.nextInt(), k = in.nextInt();
        int[] fr = new int[n - 1], to = new int[n - 1];
        for (int i = 0; i < n - 1; i++) {
            fr[i] = in.nextInt() - 1;
            to[i] = in.nextInt() - 1;
        }
        G = build_graph(n, fr, to);
        
        int[][] ret = bfs(G, 0);
        par = ret[0]; dep = ret[1];
        int s = 0;
        for (int i = 0; i < n; i++) if (dep[i] > dep[s]) s = i;
        
        ret = bfs(G, s);
        par = ret[0]; dep = ret[1];
        int t = s;
        for (int i = 0; i < n; i++) if (dep[i] > dep[t]) t = i;
        
        int dia = dep[t];
        if (dia != k * 2) {
            out.println("No");
            return;
        }
        
        int c = t;
        for (int i = 0; i < k; i++) {
            c = par[c];
        }
        
        ret = bfs(G, c);
        par = ret[0]; dep = ret[1];
        for (int i = 0; i < n; i++) {
            if (i == c) {
                if (G[i].length < 3) {
                    out.println("No");
                    return;
                }
                continue;
            }
            if (G[i].length == 1) {
                if (dep[i] != k) {
                    out.println("No");
                    return;
                }
                continue;
            }
            if (G[i].length < 4) {
                out.println("No");
                return;
            }
        }
        out.println("Yes");
    }
    
    int[][] bfs(int[][] G, int root) {
        int n = G.length;
        
        int[] par = new int[n];
        Arrays.fill(par, -1);
        
        int[] dep = new int[n];
        dep[root] = 0;
        
        int[] qu = new int[n];
        qu[0] = root;
        for (int l = 0, r = 1; l < r; l++) {
            int u = qu[l];
            for (int v : G[u]) {
                if (v != par[u]) {
                    qu[r++] = v;
                    par[v] = u;
                    dep[v] = dep[u] + 1;
                }
            }
        }
        
        return new int[][]{par, dep, qu};
    }
    
    int[][] build_graph(int n, int[] from, int[] to) {
        int[][] G = new int[n][];
        int[] cnt = new int[n];
        for (int i = 0; i < from.length; i++) {
            cnt[from[i]]++;
            cnt[to[i]]++;
        }
        for (int i = 0; i < n; i++) G[i] = new int[cnt[i]];
        for (int i = 0; i < from.length; i++) {
            G[from[i]][--cnt[from[i]]] = to[i];
            G[to[i]][--cnt[to[i]]] = from[i];
        }
        return G;
    }
    
    public static void main(String[] args) {
        in = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(System.out);
        new Multihedgehog().solve();
        out.close();
    }
    
    static FastScanner in;
    static PrintWriter out;
    
    static class FastScanner {
        BufferedReader in;
        StringTokenizer st;
        
        public FastScanner(BufferedReader in) {
            this.in = in;
        }
        
        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(in.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        
        public long nextLong() {
            return Long.parseLong(nextToken());
        }
        
        public double nextDouble() {
            return Double.parseDouble(nextToken());
        }
    }
}
