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>
using namespace std;

#define ll long long
#define endl '\n'

const int mx=2e5+5;
set<int> g[mx];
int nod[mx],n;
int F[30]={1,1},fid=2;
void dfs(int u=1,int par=-1){
	nod[u]=1;
	for(auto v:g[u]){
		if(v!=par){
			dfs(v,u);
			nod[u]+=nod[v];
		}
	}
}
int ID1=-1,ID2=-1,U,V,ok=0;
void reroot(int u,int par,int id){
	for(auto v:g[u]){
		if(v!=par){
			int a=nod[u],b=nod[v];
			nod[u]-=nod[v];
			if(nod[v]==F[id-1]){
				V=v,U=u,ID1=id-1,ID2=id-2,ok=1;
				return;
			}else if(nod[v]==F[id-2]){
				V=v,U=u,ID1=id-2,ID2=id-1,ok=1;
				return;
			}
			nod[v]+=nod[u];
			reroot(v,u,id);
			if(ok)return;
			nod[v]-=nod[u];
			nod[u]+=nod[v];
		}
	}
}
int yo(int u,int v,int id){
	if(id<0)return 0;
	if(id<2)return nod[u]==1;
	ID1=ID2=-1; ok=0;
	reroot(u,v,id);
	int a=ID1,b=ID2,c=U,d=V;
	g[c].erase(d); g[d].erase(c);
	int ans=yo(d,-1,a)&&yo(c,-1,b);
	return ans;
}

int32_t main(){
	ios_base::sync_with_stdio(0); cin.tie(0);
	for(int i=2;;i++){
		F[fid++]=F[i-1]+F[i-2];
		if(F[i]>=mx)break;
	}
	cin>>n;
	for(int i=1,u,v;i<n;i++){
		cin>>u>>v;
		g[u].insert(v);
		g[v].insert(u);
	}
	dfs();
	int id=-1;
	for(int i=0;i<fid;i++)if(F[i]==n){
		id=i;
	}
	if(id==-1)cout<<"NO\n";
	else {
		int a=yo(1,-1,id);
		if(a)cout<<"YES\n";
		else cout<<"NO\n";
	}
}