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 <bits/stdc++.h>
using namespace std;
#define intt long long int
#define F first 
#define S second
#define pb push_back
const int M=200007;
set<int> con[M];
int fib[30];
int siz[M]={0};
int dfs(int,int,int,int,int);
int check(int st,int pp,int qq)
{
    if(qq<=3) return 1;
    int aa,bb;
    int ind=lower_bound(fib,fib+30,qq)-fib;
    
    if(ind==30 || fib[ind]!=qq) 
    return -1;
    aa=fib[ind-2];
    bb=fib[ind-1];
   return dfs(st,pp,aa,bb,qq);
}
int dfs(int st,int pp,int aa,int bb,int qq)
{
    siz[st]=1;
    for(auto i:con[st])
    {
        int ss=i;
        if(ss==pp) continue;
        int ch=dfs(ss,st,aa,bb,qq);
        if(ch==1) return 1;
        else if(ch==-1) return -1; 
        int sizee=siz[ss];
        if(sizee==aa || sizee==bb)
        {
           con[st].erase(ss);
           con[ss].erase(st);
            
           int cc1=check(ss,st,sizee);
           int cc2=check(st,ss,qq-sizee);
            if(cc1==1 && cc2==1 ) return 1;
            else return -1;// for surely no answer
        }
        siz[st]+=siz[ss];
    }
    return 0;
}
void func()
{
fib[0]=fib[1]=1;
for(int i=2;i<30;i++) fib[i]=fib[i-1]+fib[i-2];
}

int main()
{
    func();
    int n;cin>>n;
    for(int i=0;i<n-1;i++)
    {
        int x,y;cin>>x>>y;
        con[x].insert(y);
        con[y].insert(x);
    }
   
    if(check(1,-1,n)==1) cout<<"YES"<<endl;
    else cout<<"NO"<<endl;
    return 0;
}