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;

vector<int> F;

vector<vector<int>> edges;
vector<int> treesize;
vector<int> parent;

void calc_size(int p, int n) {
	parent[n] = p;
	for (auto e : edges[n]) {
		if (e == p) continue;
		calc_size(n, e);
	}
	
	treesize[n] = 1;
	for (auto e : edges[n]) {
		if (e == p) continue;
		treesize[n] += treesize[e];
	}
}

int find_size(int p, int n, int target) {
	if (target == treesize[n]) return n;
	if (target > treesize[n]) return -1;
	
	for (auto e : edges[n]) {
		if (e == p) continue;
		int r = find_size(n, e, target);
		if (r >= 0) return r;
	}
	return -1;
}

int main() {
	int n;
	cin >> n;
	
	edges.resize(n);
	parent.resize(n);
	treesize.resize(n);
	
	for (int i=1; i<n; i++) {
		int u, v;
		cin >> u >> v;
		u--;
		v--;
		edges[u].push_back(v);
		edges[v].push_back(u);
	}
	
	// Precalc fib numbers
	F.push_back(1);
	F.push_back(1);
	while (F.back() < n) {
		F.push_back(F[F.size()-2]+F[F.size()-1]);
	}
	
	// Check global tree size
	if (F.back() != n) {
		cout << "NO" << endl;
		return 0;
	}
	
	queue<pair<int, int>> work;
	work.push({0, F.size()-1});
	while (!work.empty()) {
		auto cur = work.front();
		work.pop();
		if (F[cur.second] <= 2) continue;
		
		calc_size(-1, cur.first);
		int r1 = find_size(-1, cur.first, F[cur.second-1]);
		if (r1 >= 0) {
			for (auto &e : edges[r1]) {
				if (e == parent[r1])
					swap(e, edges[r1].back());
			}
			for (auto &e : edges[parent[r1]]) {
				if (e == r1)
					swap(e, edges[parent[r1]].back());
			}
			edges[r1].pop_back();
			edges[parent[r1]].pop_back();
			
			work.push({r1, cur.second-1});
			work.push({cur.first, cur.second-2});
			continue;
		}
		int r2 = find_size(-1, cur.first, F[cur.second-2]);
		if (r2 == -1) {
			cout << "NO" << endl;
			return 0;
		}
		
		for (auto &e : edges[r2]) {
			if (e == parent[r2])
				swap(e, edges[r2].back());
		}
		for (auto &e : edges[parent[r2]]) {
			if (e == r2)
				swap(e, edges[parent[r2]].back());
		}
		edges[r2].pop_back();
		edges[parent[r2]].pop_back();
		
		work.push({r2, cur.second-2});
		work.push({cur.first, cur.second-1});
	}
	
	cout << "YES" << endl;
}