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 
#define inf 0x3f3f3f3f
#define all(x) (x).begin(),(x).end()
#define P pair<int,int>
using namespace std;
const int maxn=2e5+5;
vector<int>lik[maxn];
int n;
bool fib[maxn];
P f[maxn];
void init()
{
    fib[1]=1;
    int f1=1,f2=1;
    while(1) 
    {
        int f3=f1+f2;
        if(f3>=maxn) break;
        fib[f3]=1;
        f[f3]=P(f1,f2);
        f1=f2;f2=f3;
    }
    scanf("%d",&n);
    int u,v;
    for(int i=1;i<n;++i) 
    {
        scanf("%d%d",&u,&v);
        lik[u].push_back(v);
        lik[v].push_back(u);
    }
}
int sz[maxn],fa[maxn];
int p,dn;
int xx;
void dfs1(int u,int faa,int tar)
{
    sz[u]=1;
    fa[u]=faa;
    for(auto v:lik[u])
    {
        if(v!=faa) 
        {
            dfs1(v,u,tar);
            sz[u]+=sz[v];
        }
    }
    if(sz[u]==f[tar].first||sz[u]==f[tar].second) p=u,dn=sz[u];
}
void sol()
{
    if(!fib[n]) {puts("NO");return;}
    queue<P>q;
    if(n!=1) q.push(P(1,n));
    while(!q.empty()) 
    {
        P u=q.front();q.pop();
        p=dn=0;
        dfs1(u.first,fa[u.first],u.second);
        if(p==0) {puts("NO");return;}
        int tmp=fa[p];
        for(int i=0;i<lik[tmp].size();++i)
        {
            if(lik[tmp][i]==p) lik[tmp][i]=fa[tmp];
        }
        if(dn!=1) q.push(P(p,dn));
        if(u.second-dn!=1) q.push(P(u.first,u.second-dn));
    }
    puts("YES");
}
int main()
{
    //freopen("in.txt","r",stdin);
    int T;
    //for(scanf("%d",&T);T;T--)
    {
        init();
        sol();
    }
    //cout<<xx;
    return 0;
}