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 ff first
#define ss second
#define pb push_back
#define int long long
#define ld long double
#define f(i,x,n) for(int i=x;i<n;i++)
using namespace std;
const int mod=1e9+7;
const int inf=1e18;
set <int> gp[200005];
vector <int> fib;
int dp[200005][28],ans=0;
pair <int,int> dfs(int root, int par, int no) {
    // if(par==0)
    //     cout<<root<<" "<<par<<" "<<no<<'\n';
    if(no<=3) {
        return {0,0};
    }
    pair <int,int> ret={0,1};
    for(auto i: gp[root]) {
        if(i!=par) {
            pair <int,int> curr=dfs(i,root,no);
                // cout<<"Idhar\n";
            if(curr.ss==fib[no-1] || curr.ss==fib[no-2]) {
                gp[root].erase(i);
                gp[i].erase(root);
                if(curr.ss==fib[no-1]) {
                    dfs(i,0,no-1);
                    dfs(root,0,no-2);
                }
                else {
                    dfs(i,0,no-2);
                    dfs(root,0,no-1);
                }
                return {1,0};
            }
            if(curr.ff==1)
                return {1,0};
            ret.ff+=curr.ff;
            ret.ss+=curr.ss;
        }
    }
    if(par==0) {
        ans=-1;
    }
    return ret;
}
void solve() {
    int n;
    cin>>n;
    fib.push_back(1);
    fib.push_back(1);
    int fibsz=2;
    while(fib[fibsz-1]+fib[fibsz-2]<=n) {
        fib.push_back(fib[fibsz-1]+fib[fibsz-2]);
        fibsz++;
    }
    f(i,0,n-1) {
        int u,v;
        cin>>u>>v;
        gp[u].insert(v);
        gp[v].insert(u);
    }

    if(fib[fibsz-1]!=n) {
        ans=-1;
    }
    else {
        dfs(1,0,fibsz-1);
    }
    if(ans==-1) {
        cout<<"NO\n";
    }
    else {
        cout<<"YES\n";
    }
}

int32_t main() {
    ios_base::sync_with_stdio(false); cin.tie(NULL);
    int t=1;
    // cin>>t;
    while(t--) {
        solve();
    }
    return 0;
}