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 INF 1e9
using namespace std;
typedef long long ll;
typedef pair<int,int> pi;
vector<int> fibu;
bool ans;
vector<vector<int> > gr;
vector<int> tam;
void ver_fibu();
int n;
void dfs(const int u,pi &par,const int a,const int b,const int pad){
	for(int v : gr[u])
		if(v != pad)
			dfs(v,par,a,b,u);
	tam[u] = 0;
	for(int v : gr[u]){
		if(v != pad){
			if(tam[v] == a)
				par={v,u};
			if(tam[v] == b)
				par = {u,v};
			tam[u]+=tam[v];
		}
	}
	tam[u]++;
}
void dividir(const int a,const int b){
	int x;
	for(int i = 0; i < gr[a].size(); i++)
		if(gr[a][i] == b){
			x = i;
			break;
		}
	if(gr[a].size() > 1 && x+1 != gr[a].size())
		swap(gr[a][x],gr[a][gr[a].size()-1]);
	gr[a].pop_back();
	for(int i = 0; i < gr[b].size(); i++)
		if(gr[b][i] == a){
			x = i;
			break;
		}
	if(gr[b].size() > 1 && x+1 != gr[b].size())
		swap(gr[b][x],gr[b][gr[b].size()-1]);
	gr[b].pop_back();
}
void solve(const int k,const int u){
	if(k >= 2){
		//printf("entra solve:%d\n", k);
		pi par={-1,-1};
		//tam.assign(n+1,0);
		dfs(u,par,fibu[k-1],fibu[k-2],-1);
		//printf("luego dfs\n");
		if(par.first == -1)
			ans = 0;
		else{
			dividir(par.first,par.second);
			solve(k-1,par.first);
			solve(k-2,par.second);
		}
	}
	//return 1;
}
int main(){
	ver_fibu();
	scanf("%d",&n);
	gr.assign(n+1,vector<int>());
	tam.assign(n+1,0);
	for(int i = 0; i < n-1; i++){
		int a,b;
		scanf("%d %d",&a,&b);
		gr[a].push_back(b);
		gr[b].push_back(a);
	}
	ans = 1;
	int t = lower_bound(fibu.begin(), fibu.end(), n) - fibu.begin();
	if(fibu[t] == n)
		solve(t,1);
	else
		ans = 0;
	if(ans)
		printf("YES\n");
	else
		printf("NO\n");
	return 0;
}

void ver_fibu(){
	int lim = 200000;
	int x = 2;
	int i = 2;
	fibu.reserve(28);
	fibu.push_back(1);
	fibu.push_back(1);
	while(x <= lim){
		fibu.push_back(x);
		x = fibu[i] + fibu[i-1];
		i++;
	}
	fibu.push_back(INF);
/*	printf("fibu(%d): ",(int)v.size());
	for(int ff : v)
		printf("%d ", ff);
	printf("\n");*/
}