#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 500;
queue<int> q;
int n, deg[N], ty[N], bio[N], cnt = 0, k, rl[N];
vector<int> v[N];
int main() {
  memset(ty, -1, sizeof(ty));
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    deg[x]++, deg[y]++;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (int i = 1; i <= n; i++)
    if (deg[i] == 1) q.push(i), ty[i] = 0;
  while (!q.empty()) {
    cnt++;
    int x = q.front();
    q.pop();
    bio[x] = 1;
    if (ty[x] && rl[x] < 3) {
      printf("No\n");
      return 0;
    }
    if (cnt == n) {
      if (ty[x] == k) {
        printf("Yes\n");
        return 0;
      }
      break;
    }
    if (deg[x] != 1) {
      printf("No\n");
      return 0;
    }
    for (int y : v[x]) {
      if (bio[y]) continue;
      deg[y]--;
      rl[y]++;
      if (ty[y] != -1 && ty[y] != ty[x] + 1) {
        printf("No\n");
        return 0;
      }
      ty[y] = ty[x] + 1;
      if (deg[y] == 1) q.push(y);
    }
  }
  printf("No\n");
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 500;
queue<int> q;
int n, deg[N], ty[N], bio[N], cnt = 0, k, rl[N];
vector<int> v[N];
int main() {
  memset(ty, -1, sizeof(ty));
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    deg[x]++, deg[y]++;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (int i = 1; i <= n; i++)
    if (deg[i] == 1) q.push(i), ty[i] = 0;
  while (!q.empty()) {
    cnt++;
    int x = q.front();
    q.pop();
    bio[x] = 1;
    if (ty[x] && rl[x] < 3) {
      printf("No\n");
      return 0;
    }
    if (cnt == n) {
      if (ty[x] == k) {
        printf("Yes\n");
        return 0;
      }
      break;
    }
    if (deg[x] != 1) {
      printf("No\n");
      return 0;
    }
    for (int y : v[x]) {
      if (bio[y]) continue;
      deg[y]--;
      rl[y]++;
      if (ty[y] != -1 && ty[y] != ty[x] + 1) {
        printf("No\n");
        return 0;
      }
      ty[y] = ty[x] + 1;
      if (deg[y] == 1) q.push(y);
    }
  }
  printf("No\n");
  return 0;
}

```