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 int long long
#define ld long double
#define vc vector
#define s second
#define f first

#define p(v) cout << #v << ": " << v << endl;
#define print(v) cout << #v << ": "; for (auto x : v )cout << x << " "; cout << endl;


//#include 'prettyprint.hpp'

using namespace std;


typedef pair <int, int> pii;


int xmax(int & a, int b){
    return a = max(a, b);
}
int xmin(int & a, int b){
    return a = min(a, b);
}

vc <int> fib;



pair <int, int> split;

int dfs (vc <unordered_set<int > >& adi, int node, int k, int par){
    int sum = 1;
    for (auto next: adi[node]){
        if (next == par) continue;
        int add = dfs(adi, next, k, node);
        if (add == fib[k-1]){
            split = {node, next};
        }
        if (add == fib[k-2]){
            split = {next, node};
        }
        sum+=add;
    }
    return sum;

}


bool fib_split(vc <unordered_set<int > > & adi, int start, int k){
    //cout << "starting " << start << " " << k << endl;
    if (k==0 ||  k == 1) return true;
    split = {-1, -1};
    dfs (adi, start, k, -1);
    //cout << split.f << " " << split.s << endl;
    if (split.f == -1) return false;
    adi[split.f].erase(split.s);
    adi[split.s].erase(split.f);
    //cerr << split.f << " " << split.s << endl;
    //cout << "shoutl start " << "( " << split.f << " , "<< k-2 << " ) ( " << split.s << " , " << k-1 << " )" << endl;
    pii cur_s = split;
    bool g1 = fib_split(adi,cur_s.f, k-2);
    bool g2 = fib_split(adi, cur_s.s, k-1);
    //cout << "return " << start << " " << k << endl;
    return g1&&g2;
}


void do_test(){
   int n; cin >> n;
   vc <unordered_set<int> > adi (n);

   for (int i=0; i+1<n; i++){
        int u, v; cin >> u >> v; u--; v--;
        adi[u].insert(v);
        adi[v].insert(u);
    }
    fib.push_back(1);
    fib.push_back(1);
    while(fib.back()<n){
        fib.push_back(fib[fib.size()-1] + fib[fib.size()-2]);
    }
    if (fib.back()!=n){
        cout << "NO\n";
        return;
    }
    bool geht = fib_split(adi, 0, fib.size()-1);
    if (geht){
        cout << "YES\n";
    }else{
        cout << "NO\n";
    }

}


signed main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    do_test();
}