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;
typedef double db;
typedef long long ll;
typedef pair<int, int> pii;
#define fi first
#define se second
#define empb emplace_back
#define all(x) (x).begin(),(x).end()



int main() {
#ifdef local
	freopen("in.txt", "r", stdin);
#endif
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	int n;
	cin >> n;
	vector<pii> edges;
	vector<vector<int> > E(n);
	for(int i = 1, u, v; i < n; i++) {
		cin >> u >> v;
		--u, --v;
		E[u].push_back(edges.size());
		edges.emplace_back(u, v);
		E[v].push_back(edges.size());
		edges.emplace_back(v, u);
	}
	if(n <= 3) {
		cout << "YES\n";
		return 0;
	}
	vector<int> fib = {1, 2};
	int u = 2, v = 1;
	while(u < n) {
		int t = u;
		u = u + v;
		v = t;
		fib.push_back(u);
	}
	if(fib.back() != n) {
		cout << "NO\n";
		return 0;
	}
	vector<int> sz(n);
	vector<bool> used(2 * n - 2);
	function<void(int, int)> prepare = [&](int u, int p) {
		sz[u] = 1;
		for(auto &i : E[u]) {
			int v = edges[i].se;
			if(used[i] || v == p) continue;
			prepare(v, u);
			sz[u] += sz[v];
		}
	};
	function<int(int, int, int)> Find = [&](int u, int p, int d) {
		for(auto &i : E[u]) {
			int v = edges[i].se;
			if(used[i] || v == p) continue;
			if(sz[v] == fib[d - 1] || sz[v] == fib[d - 2]) {
				return sz[v] == fib[d - 1] ? (i ^ 1) : i;
			}
			auto r = Find(v, u, d);
			if(r != -1) {
				return r;
			}
		}
		return -1;
	};
	function<bool(int, int)> ok = [&](int u, int d) {
		if(d <= 2) {
			return true;
		}
		prepare(u, -1);
		int id = Find(u, -1, d);
		if (id == -1) {
			return false;
		}
		int x, y;
		tie(x, y) = edges[id];
		used[id] = used[id ^ 1] = 1;
		if(ok(x, d - 1) && ok(y, d - 2)) {
			return true;
		}
		return false;
	};
	int cur = fib.size() - 1;
	if(ok(0, cur)) {
		cout << "YES\n";
	} else {
		cout << "NO\n";
	}
	return 0;
}