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;
template <class T>
void chmin(T& a, const T& b) {
  if (a > b) a = b;
}
template <class T>
void chmax(T& a, const T& b) {
  if (a < b) a = b;
}
int n, k;
vector<int> g[100100];
int deg[100100];
int par[100100];
int depth[100100];
vector<int> vec[100100];
bool key = false;
int max_depth = -1;
int dv = -1;
void dfs(int v, int p, int d) {
  depth[v] = d;
  par[v] = p;
  if (max_depth < d) {
    max_depth = d;
    dv = v;
  }
  if (key) vec[d].push_back(v);
  for (int i = 0; i < g[v].size(); i++) {
    int u = g[v][i];
    if (u == p) continue;
    dfs(u, v, d + 1);
  }
}
int main() {
  scanf("%d %d", &n, &k);
  int s = -1;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
    deg[u]++;
    deg[v]++;
  }
  for (int i = 0; i < n; i++)
    if (deg[i] == 1) s = i;
  if (s == -1) {
    printf("No\n");
    return 0;
  }
  dfs(s, -1, 0);
  if (max_depth % 2 != 0) {
    printf("No\n");
    return 0;
  }
  int hd = max_depth / 2;
  for (int i = 0; i < hd; i++) dv = par[dv];
  int center = dv;
  if (deg[center] < 3) {
    printf("No\n");
    return 0;
  }
  key = true;
  dfs(center, -1, 0);
  for (int i = 1; i < n; i++) {
    if ((i > k && vec[i].size() != 0) || (i <= k && vec[i].size() == 0)) {
      printf("No\n");
      return 0;
    }
    if (vec[i].size() == 0) continue;
    bool flag = true;
    bool one = (deg[vec[i][0]] == 1) ? true : false;
    for (int j = 0; j < vec[i].size(); j++) {
      if (one && deg[vec[i][j]] != 1) {
        flag = false;
        break;
      }
      if (!one && deg[vec[i][j]] <= 3) {
        flag = false;
        break;
      }
    }
    if (!flag) {
      printf("No\n");
      return 0;
    }
  }
  printf("Yes\n");
  return 0;
}
