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:
// Problem: E. Fib-tree
// Contest: Codeforces - Codeforces Global Round 13
// URL: https://codeforc.es/contest/1491/problem/E
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
#define rep(i,x,y) for(auto i=(x);i<=(y);++i)
#define dep(i,x,y) for(auto i=(x);i>=(y);--i)
#define fr first
#define sc second
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,ll>pil;
typedef pair<int,int>pii;
typedef pair<ll,ll>pll;
const int N=1e6+10;
const int inf=1e9;
const int mo=998244353;
ll a[N]={1,1};
map<ll,int> mp;
set<int> G[N];
//想了想好像直接切就行，又要假的错觉 //....原来真的是直接切就行啊  
int p,f,y,z;
bool b;
int dfs(int u,int fa){
    int res=1;
    for(auto v:G[u]){
        if(v==fa) continue;
        res+=dfs(v,u);
    }
    if(fa!=-1&&res==y) p=u,f=fa,b=0;
    if(fa!=-1&&res==z) p=u,f=fa,b=1;
    return res;
}
bool check(int u,int n){
    if(n==1) return 1;
    if(!mp.count(n)) return 0;
    p=f=0;
    y=a[mp[n]-2],z=a[mp[n]-1];
    dfs(u,-1);
    if(!p) return 0;
    else {
        G[p].erase(f);
        G[f].erase(p);
        int pp=p,ff=f,yy=y,zz=z; //确实zz*2  zz本z
        if(!b) return check(pp,yy)&&check(ff,zz);
        else return check(pp,zz)&&check(ff,yy);
    }
}
int main(){
    //freopen("D:\\in.txt","r",stdin);
    //freopen("D:\\out.txt","w",stdout);
    rep(i,2,60) a[i]=a[i-1]+a[i-2];
    rep(i,1,60) mp[a[i]]=i;
    int n; cin>>n;
    rep(i,1,n-1){
        int u,v; cin>>u>>v;
        G[u].insert(v);
        G[v].insert(u);
    }
    if(check(1,n)) cout<<"YES\n";
    else cout<<"NO\n";
    return 0;
}