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:
///Efectiv de ce intra acest lucru in timp

#include <iostream>
#include <vector>
#include <map>
#pragma GCC optimize("O3")
#define pi std::pair<int, int>

const int nmax = 4e5 + 5;

std::vector<int>fibo={1, 1};
std::vector<bool>isf(nmax, 0);
std::vector<int>l[nmax];
std::map<pi, int>nono;
int dim[nmax][100];

bool solve(int node, int last, int sz, int dimcst) {
	if(sz==1) return true;
	if(!isf[sz]) return false;
	dim[node][dimcst] = 1;
	for(int next:l[node]) if(last!=next and nono.find({node, next})==nono.end() and nono.find({next, node})==nono.end()) {
		if(solve(next, node, sz, dimcst)) return true;
		dim[node][dimcst] += dim[next][dimcst];
	}
	int x = dim[node][dimcst];
	if(!isf[x] or !isf[sz-x]) return false;
	nono[{node, last}] = 1;
	if(solve(node, -1, x, dimcst + 1) and solve(last, -1, sz - x, dimcst+1)) return true;
	nono.erase({node, last});
	return false;
}

int main() {
	std::ios::sync_with_stdio(false);
	std::cin.tie(0);
	std::cout.tie(0);
	int n, u, v;
	std::cin>>n;
	isf[1] = true;
	while(fibo.back()<=n) fibo.push_back(fibo[fibo.size()-2] + fibo[fibo.size()-1]), isf[fibo.back()]=true;
	fibo.pop_back();
	for(int i=1;i<n;i++) {
		std::cin>>u>>v;
		l[u].push_back(v);
		l[v].push_back(u);
	}
	if(fibo.back()!=n) {
		std::cout<<"No";
		return 0;
	}
	std::cout<<((solve(1, -1, n, 0))?"Yes":"No");
}