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 <set>
#include <map>
#include <cmath>
#include <queue>
#include <vector>
#include <bitset>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
template <class T> bool umin(T &x,T y) {return y<x? x=y,1: 0;}
template <class T> bool umax(T &x,T y) {return y>x? x=y,1: 0;}
const int N = 2e5+5;
vector<int> G[N];
int n,fib[N],m,siz[N],a[N],cnt=0;

int tot = 0, col[N];
void prp(int u,int fa,int st) {
	++cnt; a[st+cnt] = u; siz[u] = 1;
	// cout << u << " " << fa << "\n";
	for(int i=0;i<(int)G[u].size();++i) {
		int v = G[u][i]; 
		if(v == fa || col[v] != col[u]) continue;
		prp(v,u,st);
		siz[u] += siz[v];
	}
}

bool check(int l,int r,int m) {
	if(l == r) return 1;
	for(int i=l;i<=r;++i) {
		if(siz[a[i]] == fib[m-1] || siz[a[i]] == fib[m-2]) {

			tot++; for(int j=i;j<=i+siz[a[i]]-1;++j) col[a[j]] = tot;
			tot++; int beg;
			for(int j=l;j<=r;++j) if(col[a[j]] != tot-1) {
				col[a[j]] = tot;
				beg = a[j];
			}
			// if(m == 4) {
			// 	cout << a[i] << "?\n";
			// 	for(int i=l;i<=r;++i) cout << a[i] << " "; cout << "\n";
			// 	for(int i=l;i<=r;++i) cout << col[a[i]] << " "; cout << "\n";
			// }

			cnt = 0;
			int ai = a[i], sz = siz[a[i]];
			prp(a[i],0,l-1);
			prp(beg,0,l-1);

			int t1,t2;
			if(sz == fib[m-1]) t1 = m-1, t2 = m-2;
			else t1 = m-2, t2 = m-1;
			return check(l,l+sz-1,t1) && check(l+sz,r,t2); 
		}
	}

	return 0;
}

signed main()
{
	ios::sync_with_stdio(false);
	cin >> n;

	if(n == 1) {
		puts("YES");
		return 0;
	}
	fib[0]=fib[1]=1;
	for(m=2;;++m) {
		fib[m] = fib[m-1]+fib[m-2];
		if(fib[m] >= n) break;
	}

	if(fib[m] != n) {
		puts("NO");
		return 0;
	}
	// cout << m << "\n";

	for(int i=1,u,v;i<n;++i) {
		cin >> u >> v;
		G[u].push_back(v);
		G[v].push_back(u);
	}

	memset(col,0,sizeof(col));
	cnt = 0; prp(1,0,0);
	// for(int i=1;i<=n;++i) cout << a[i] << " "; cout << "\n";
	// for(int i=1;i<=n;++i) cout << siz[a[i]] << " "; cout << "\n";

	if(check(1,n,m)) puts("YES");
	else puts("NO");

	return 0;
}