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>
using namespace std;
#define int long long
typedef pair<int,int> pi;
#define f first
#define s second
#define FAST ios_base::sync_with_stdio(0); cin.tie(0);
#define all(x) x.begin(),x.end()
typedef pair<int, pi> pii;

const int maxn = 200010;
const int INF = LLONG_MAX/2;

int fib[40];
set <int> adjlist[maxn];
map <int, int> mp;
int sz[maxn];

pi curans;

void dfs(int x, int p, int total) {
	sz[x] = 1;
	for (auto i : adjlist[x]) if (i != p) {
		dfs(i,x,total);
		sz[x] += sz[i];
	}
	if (mp.find(sz[x]) != mp.end() and mp.find(total-sz[x]) != mp.end()) curans = pi(x,p);
}

int32_t main() {
	FAST
	int n; cin >> n;
	for (int i =0;i<n-1;i++) {
		int a,b; cin >> a >> b;
		adjlist[a].insert(b);
		adjlist[b].insert(a);
	}
	
	fib[0] = fib[1] = 1;
	mp[1] = 1;
	for (int i =2;i<40;i++) {
		fib[i] = fib[i-1] + fib[i-2];
		mp[fib[i]] = 1;
	}
	
	if (mp.find(n) == mp.end()) {
		cout << "NO";
		return 0;
	}
	
	vector <pi> v;
	v.push_back(pi(1,n));
	
	while (true) {
		vector <pi> vv;
		for (auto i: v) {
			if (i.s == 1) continue;
			curans = pi(-1,-1);
			
			dfs(i.f,-1,i.s);
			auto [x,p] = curans;
			if (x == -1) {
				cout << "NO";
				return 0;
			}
			
			adjlist[x].erase(p);
			adjlist[p].erase(x);
			
			vv.push_back(pi(x,sz[x]));
			vv.push_back(pi(p,i.s - sz[x]));
		}
		v.clear();
		v = vv;
		if (v.empty()) break;
	}
	cout << "YES";
}