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>
inline int lowbit(int x) { return x & -x; }
const int N = 1e5 + 10;
const int mod = (int)998244353;
const int INF = 0x3f3f3f3f;
const long long LINF = (1LL << 62);
const double PI = acos(-1.0);
using namespace std;
int n, k;
int deg[N], head[N], cnt, tag[N];
int treeDiameter, point, tagPoint;
struct Edge {
  int nex, to;
} edge[N << 1];
inline void AddEdge(int u, int v) {
  edge[++cnt].nex = head[u];
  edge[cnt].to = v;
  head[u] = cnt;
  edge[++cnt].nex = head[v];
  edge[cnt].to = u;
  head[v] = cnt;
}
void dfs(int u, int fa, int len) {
  if (len > treeDiameter) {
    treeDiameter = len;
    point = u;
  }
  for (int i = head[u]; i; i = edge[i].nex) {
    int v = edge[i].to;
    if (v != fa) {
      dfs(v, u, len + 1);
    }
  }
}
bool dfs1(int u, int fa, int len) {
  if (len == treeDiameter) {
    return false;
  }
  for (int i = head[u]; i; i = edge[i].nex) {
    int v = edge[i].to;
    if (v != fa) {
      if (!dfs1(v, u, len + 1)) {
        if (len == (treeDiameter >> 1)) {
          tagPoint = u;
        }
        return false;
      }
    }
  }
  return true;
}
bool dfs2(int u, int fa, int len) {
  if (fa && deg[u] == 1) {
    if (len == (treeDiameter >> 1))
      return true;
    else
      return false;
  }
  int cnt = 0;
  for (int i = head[u]; i; i = edge[i].nex) {
    int v = edge[i].to;
    if (v != fa) {
      if (!dfs2(v, u, len + 1)) return false;
      cnt++;
    }
  }
  if (cnt >= 3)
    return true;
  else
    return false;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    deg[u]++;
    deg[v]++;
    AddEdge(u, v);
  }
  for (int i = 1; i <= n; i++)
    if (deg[i] == 1) {
      tag[edge[head[i]].to]++;
    }
  for (int i = 1; i <= n; i++)
    if (tag[i] && tag[i] < 3) {
      printf("No\n");
      return 0;
    }
  dfs(1, 0, 0);
  dfs(point, 0, 0);
  if ((treeDiameter & 1) || k != (treeDiameter >> 1)) {
    printf("No\n");
    return 0;
  }
  dfs1(point, 0, 0);
  if (dfs2(tagPoint, 0, 0))
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}
