#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
vector<vector<int> > g(100005);
int to, maxD, center;
int n, k;
void dfs(int u, int p = -1, int d = 0) {
  if (d > maxD) {
    to = u;
    maxD = d;
  }
  for (auto v : g[u]) {
    if (v != p) {
      dfs(v, u, d + 1);
    }
  }
}
vector<int> path;
void dfs2(int u, int p = -1) {
  path.push_back(u);
  if (u == to) {
    center = path[path.size() / 2];
  }
  for (auto v : g[u]) {
    if (v != p) dfs2(v, u);
  }
  path.pop_back();
}
void dfs3(int u, int p = -1, int depth = 0) {
  if (depth == k + 1) {
    cout << "No";
    exit(0);
  }
  int cnt = 0;
  for (auto v : g[u]) {
    if (v == p) continue;
    cnt++;
    dfs3(v, u, depth + 1);
  }
  if (cnt < 3 && depth != k) {
    cout << "No";
    exit(0);
  }
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0);
  maxD = 0;
  int from = to;
  dfs(to);
  if (maxD != 2 * k) {
    cout << "No";
    return 0;
  }
  dfs2(from);
  dfs3(center);
  cout << "Yes" << endl;
}

### Prompt

In Cpp, your task is to solve 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 mod = 998244353;
vector<vector<int> > g(100005);
int to, maxD, center;
int n, k;
void dfs(int u, int p = -1, int d = 0) {
  if (d > maxD) {
    to = u;
    maxD = d;
  }
  for (auto v : g[u]) {
    if (v != p) {
      dfs(v, u, d + 1);
    }
  }
}
vector<int> path;
void dfs2(int u, int p = -1) {
  path.push_back(u);
  if (u == to) {
    center = path[path.size() / 2];
  }
  for (auto v : g[u]) {
    if (v != p) dfs2(v, u);
  }
  path.pop_back();
}
void dfs3(int u, int p = -1, int depth = 0) {
  if (depth == k + 1) {
    cout << "No";
    exit(0);
  }
  int cnt = 0;
  for (auto v : g[u]) {
    if (v == p) continue;
    cnt++;
    dfs3(v, u, depth + 1);
  }
  if (cnt < 3 && depth != k) {
    cout << "No";
    exit(0);
  }
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0);
  maxD = 0;
  int from = to;
  dfs(to);
  if (maxD != 2 * k) {
    cout << "No";
    return 0;
  }
  dfs2(from);
  dfs3(center);
  cout << "Yes" << endl;
}

```