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>
#define ll  long long
using namespace std;
vector<ll> f(2);
vector<ll> val;
vector<vector<ll> > h;
pair<ll,ll> dfs(ll ind,ll par,ll v){
    val[ind]=1;
    for(int i=0;i<h[ind].size();i++){
        if(h[ind][i]==par){
            h[ind][i]=h[ind][h[ind].size()-1];
            h[ind].pop_back();
            i--;
        }else{
            pair<ll,ll> r=dfs(h[ind][i],ind,v);
            if(r.first!=-1)
                return r;
            val[ind]+=val[h[ind][i]];
        }
    }
    if(val[ind]==f[v-1]||val[ind]==f[v-2])
        return make_pair(ind,par);
    else
        return make_pair(-1,par);
}
bool check(ll r,ll s){
    if(s<=3)
        return true;
    pair<ll,ll> t=dfs(r,-1,s);
    if(t.first==-1)
        return false;
    ll par=t.second;
    for(int i=0;i<h[par].size();i++)
        if(h[par][i]==t.first){
            h[par][i]=h[par][h[par].size()-1];
            h[par].pop_back();
            break;
        }
    if(val[t.first]==f[s-2])
    return (check(r,s-1)&&check(t.first,s-2));
    else
        return (check(r,s-2)&&check(t.first,s-1));
}
int main() {
    f[0]=1;
    f[1]=1;
    while(f[f.size()-1]<500009){
        f.push_back(f[f.size()-1]+f[f.size()-2]);
    }
    ll n;
    cin>>n;
    ll ind=-1;
    for(int i=0;i<f.size();i++){
        if(n==f[i]){
            ind=i;
            break;
        }
    }
    if(ind==-1){
        cout<<"NO"<<endl;
        return 0;
    }
    vector<ll> em;
    for(int i=0;i<n;i++){
        h.push_back(em);
        val.push_back(0);
    }
    ll u,v;
    for(int i=0;i<n-1;i++){
        cin>>u>>v;
        u--;
        v--;
        h[u].push_back(v);
        h[v].push_back(u);
    }
    if(check(0,ind)){
        cout<<"YES"<<endl;
    }else{
        cout<<"NO"<<endl;
    }
}