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:
/*
author: Edward Johnson(Manchester_UNITED)
problem: 1491E Fib-tree
time: 2021.03.06 04:45:49
last verdict: Wrong Answer on Test 14
*/
#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define inf 1000000000
#define inf64 1000000000000000000
#define X first
#define Y second
#define ll long long
using namespace std;
const int maxn=202020;
ll n,fibid[maxn],sz[maxn],pared[maxn],par[maxn];
bool vis[maxn],ok=true;
vector<ll> path;
vector<pair<ll,ll> > vt[maxn];
void predfs(ll x,ll ban){
	path.pb(x);
	if(x<0)
	{
		puts("WHAT THE HELL ARE YA DOING?");
		exit(0);
	}
	par[x]=ban;
	sz[x]=1;
	for(int i=0;i<vt[x].size();i++){
		if(vis[vt[x][i].Y]||vt[x][i].X==ban) continue;
		pared[vt[x][i].X]=vt[x][i].Y;
		predfs(vt[x][i].X,x);
		sz[x]+=sz[vt[x][i].X];
	}
	return;
}
void dfs(ll x,ll req){
	if(req<=4) return;
	vector<ll> tpath;
	path.clear();
	predfs(x,-1);
	tpath=path;
	for(int i=0;i<tpath.size();i++){
		if(fibid[sz[tpath[i]]]!=-1&&fibid[sz[tpath[i]]]==req-1){
			vis[pared[tpath[i]]]=true;
			dfs(tpath[i],req-1);
			dfs(x,req-2);
			return;
		}
		else if(fibid[sz[tpath[i]]]!=-1&&fibid[sz[tpath[i]]]==req-2){
			vis[pared[tpath[i]]]=true;
			dfs(tpath[i],req-2);
			dfs(x,req-1);
			return;
		}
	}
	ok=false;
	return;
}
int main(){
	memset(fibid,-1,sizeof(fibid));
	for(int i=1,a=1,b=1;a<maxn;i++){
		fibid[a]=i;
		int c=b;
		b+=a,a=c;
	}
	scanf("%lld",&n);
	for(int i=1;i<n;i++){
		ll x,y;
		scanf("%lld%lld",&x,&y);
		x--,y--;
		vt[x].pb(mp(y,i));
		vt[y].pb(mp(x,i));
	}
	if(fibid[n]==-1){
		puts("No");
		return 0;
	}
	dfs(0,fibid[n]);
	if(ok) puts("Yes");
	else puts("No");
	return 0;
}