Problem: Let F_k denote the k-th term of Fibonacci sequence, defined as below:

  * F_0 = F_1 = 1
  * for any integer n ≥ 0, F_{n+2} = F_{n+1} + F_n



You are given a tree with n vertices. Recall that a tree is a connected undirected graph without cycles.

We call a tree a Fib-tree, if its number of vertices equals F_k for some k, and at least one of the following conditions holds:

  * The tree consists of only 1 vertex;
  * You can divide it into two Fib-trees by removing some edge of the tree. 



Determine whether the given tree is a Fib-tree or not.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of vertices in the tree.

Then n-1 lines follow, each of which contains two integers u and v (1≤ u,v ≤ n, u ≠ v), representing an edge between vertices u and v. It's guaranteed that given edges form a tree.

Output

Print "YES" if the given tree is a Fib-tree, or "NO" otherwise.

You can print your answer in any case. For example, if the answer is "YES", then the output "Yes" or "yeS" will also be considered as correct answer.

Examples

Input


3
1 2
2 3


Output


YES


Input


5
1 2
1 3
1 4
1 5


Output


NO


Input


5
1 3
1 2
4 5
3 4


Output


YES

Note

In the first sample, we can cut the edge (1, 2), and the tree will be split into 2 trees of sizes 1 and 2 correspondently. Any tree of size 2 is a Fib-tree, as it can be split into 2 trees of size 1.

In the second sample, no matter what edge we cut, the tree will be split into 2 trees of sizes 1 and 4. As 4 isn't F_k for any k, it's not Fib-tree.

In the third sample, here is one possible order of cutting the edges so that all the trees in the process are Fib-trees: (1, 3), (1, 2), (4, 5), (3, 4). 

Solution (in CPP):

Solution:
#include<iostream>
#include <algorithm>
#include<vector>
using namespace std;
const int N=200000;
int p[N],next_root,next_index_in_fi;vector<int> ar[N],Fi;
void parcp(int u){
  for(int k=0;k<ar[u].size();k++){
    int v=ar[u][k];
    if(v!=p[u]){
      p[v]=u;
      parcp(v);
    }
  }
}
void remov(vector<int> v,int x){
  int k=0;
  while(v[k]!=x) k++;
  for(int i=k;i<v.size()-1;i++) v[i]=v[i+1];
  v.pop_back();
}
int parc(int u,int index_in_fi){
  int s=1;
  for(int k=0;k<ar[u].size();k++){
    int v=ar[u][k];if(v!=p[u]){
      s+=parc(v,index_in_fi);
    }
  }
  if(s==Fi[index_in_fi-1]){
    next_root=u;
    next_index_in_fi=index_in_fi-1;
  }
  if(s==Fi[index_in_fi-2]){
    next_root=u;
    next_index_in_fi=index_in_fi-2;
  }
  return s;
}
bool fi(int root,int index_in_fi){
  //cout<<root<<" "<<index_in_fi<<endl;
  //if (root==0) cout<<ar[0][4]<<endl;
  if(index_in_fi<4)
    return true;
  next_root=-1;
  parc(root,index_in_fi);
  if(next_root!=-1){
    //remov(ar[next_root],p[next_root]);
    //remov(ar[p[next_root]],next_root);
    int k=0;
    while(ar[p[next_root]][k]!=next_root) k++;
    ar[p[next_root]][k]=ar[p[next_root]][ar[p[next_root]].size()-1];
    ar[p[next_root]].pop_back();
    int nr=next_root,ni=next_index_in_fi;
    return fi(root,index_in_fi*2-3-next_index_in_fi) && fi(nr,ni);
  } return false;
}
int main(){
  int n;cin>>n;
  for(int k=0;k<n-1;k++){
    int a,b;
    cin>>a>>b;
    ar[a-1].push_back(b-1);
    ar[b-1].push_back(a-1);
  }
  Fi.push_back(1);
  Fi.push_back(1);
  p[0]=0;
  parcp(0);
  while(Fi[Fi.size()-1]<n)
    Fi.push_back(Fi[Fi.size()-1]+Fi[Fi.size()-2]);
  if(Fi[Fi.size()-1]==n && fi(0,Fi.size()-1)) cout<<"YES";
  else cout<<"NO";
}