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 debug freopen("r.txt","r",stdin)
#define mp make_pair
#define ri register int
#define pb push_back
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double lf;
typedef pair<int, int> pii;
const int maxn = 1e6+10;
const int N = 1500+10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9+7;
const int hash_num = 131;
const double eps=1e-6;
const double PI=acos(-1.0);
inline ll read(){ll s=0,w=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
return s*w;}
ll qpow(ll p,ll q){return (q&1?p:1)*(q?qpow(p*p%mod,q/2):1)%mod;}
int son[51][maxn],check[maxn],f[maxn],u,v,n;
std::vector<int> G[maxn];
set<pii> s;

int dfs(int u,int father,int num,int d)
{
	son[d][u]=1;
	if (num==1) return 1;
	if (!check[num]) return 0;
	for (auto v:G[u])
	{
		if (v==father) continue;
		if (s.count({u,v})) continue;
		if (dfs(v,u,num,d)) return 1;
		son[d][u]+=son[d][v];
	}
	if (!check[son[d][u]] || !check[num-son[d][u]]) return 0;
	s.insert({u,father});
	s.insert({father,u});
	if (dfs(u,father,son[d][u],d+1) && dfs(father,u,num-son[d][u],d+1)) return 1;
	s.erase({u,father});
	s.erase({father,u});
	return 0;
}
int main()
{
	#ifndef ONLINE_JUDGE
		debug;
	#endif
	n=read();
	f[0]=f[1]=1;
	check[1]=1;
	for (int i = 2; i <= 30; ++i)
	{
		f[i]=f[i-1]+f[i-2];
		if (f[i]<=2e5) check[f[i]]=1;
	}
	for (int i = 1; i < n ; ++i)
	{
		u=read(),v=read();
		G[u].pb(v);
		G[v].pb(u);
	}
	cout << (dfs(1,1,n,1)?"YES":"NO") <<endl;
	return 0;
}