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:
using namespace std;
#include <bits/stdc++.h>
#define N 200005
#define ll long long
int n,k;
int f[N];
struct EDGE{
	int to;
	EDGE *las;
	bool bz;
} e[N*2];
int ne;
EDGE *last[N];
void link(int u,int v){
	e[ne]={v,last[u],0};
	last[u]=e+ne++;
}
#define rev(ei) (e+(int((ei)-e)^1))
int sz[N];
void find(int x,int fa,int k,EDGE *&u,EDGE *&v,int &szu,int &szv){
	sz[x]=1;
	for (EDGE *ei=last[x];ei;ei=ei->las)
		if (ei->to!=fa && ei->bz==0){
			find(ei->to,x,k,u,v,szu,szv);
			sz[x]+=sz[ei->to];
			if (sz[ei->to]==f[k-1] || sz[ei->to]==f[k-2]){
				if (!u)
					u=ei,szu=(sz[ei->to]==f[k-1]?k-1:k-2);
				else if (!v)
					v=ei,szv=(sz[ei->to]==f[k-1]?k-1:k-2);
//				else
//					printf("fuck\n");
			}
		}
}
bool divide(int r,int k){
	if (k==0 || k==1)
		return 1;
	EDGE *u=NULL,*v=NULL;
	int szu=0,szv=0;
	find(r,0,k,u,v,szu,szv);
	if (!u) return 0;
	u->bz=rev(u)->bz=1;
	bool t=(divide(u->to,szu) && divide(r,2*k-3-szu));
	u->bz=rev(u)->bz=0;
	if (t) return 1;
	if (!v) return 0;
	v->bz=rev(v)->bz=1;
	t=(divide(v->to,szv) && divide(r,2*k-3-szv));
	v->bz=rev(v)->bz=0;
	return t;
}
//ll T[N];
int main(){
//	freopen("in.txt","r",stdin);
//	freopen("table.txt","w",stdout);
	scanf("%d",&n);
	f[0]=1,f[1]=1;
	for (k=1;f[k]<=n;++k)
		f[k+1]=f[k]+f[k-1];
	--k;
	if (f[k]!=n){
		printf("NO\n");
		return 0;
	}
//	T[0]=1,T[1]=1;
//	for (int i=2;i<=k;++i)
//		T[i]=2*(T[i-1]+T[i-2]);
//	printf("%lld\n",T[k]);
//	for (int i=1;i<=k;++i)
//		printf("%lld\n",T[i]);
//	return 0;
	for (int i=1;i<n;++i){
		int u,v;
		scanf("%d%d",&u,&v);
		link(u,v);
		link(v,u);
	}
	bool ans=divide(1,k);
	if (ans)
		printf("YES\n");
	else
		printf("NO\n");
	return 0;
}