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>
#define N 200005
using namespace std;
vector<int> fib;
int arr[N];
vector<pair<int, bool>> v[N];
int n;
void no()
{
	cout << "No";
	exit(0);
}
void calculate_size(int u, int fa)
{
	arr[u] = 1;
	for (pair<int, bool> e : v[u])
	{
		if (e.second) continue;
		int v = e.first;
		if (v == fa) continue;
		calculate_size(v, u);
		arr[u] += arr[v];
	}
}
void cut_edge(int u, int fa, int k, int& pu, int& pv, int& kd)
{
	for (pair<int, bool> e : v[u])
	{
		if (pu) return;
		if (e.second) continue;
		int v = e.first;
		if (v == fa) continue;
		if (arr[v] == fib[k - 1] || arr[v] == fib[k - 2])
		{
			pu = u; pv = v;
			kd = (arr[v] == fib[k - 1]) ? k - 1 : k - 2;
			break;
		}
		cut_edge(v, u, k, pu, pv, kd);
	}
}
void check(int u, int k)
{
	if (k <= 1) return;
	calculate_size(u, 0);
	int pu = 0, pv = 0, kd = 0;
	cut_edge(u, 0, k, pu, pv, kd);
	if (!pu) no();
	for (pair<int, bool>& e : v[pu])
		if (e.first == pv) e.second = true;
	for (pair<int, bool>& e : v[pv])
		if (e.first == pu) e.second = true;
	check(pv, kd);
	check(pu, (kd == k - 1) ? k - 2 : k - 1);
}
int main()
{
	ios::sync_with_stdio(false);
	cin >> n;
	fib.push_back(1);
	fib.push_back(1);   
	int k=1;
	while (true)
	{
		if (fib[k] >= n)
			break;
		k++;
		fib.push_back(fib[k-1] + fib[k-2]);
	}
	int a, b;
	for (int i = 1; i < n; i++)
	{
		cin >> a >> b;
		v[a].push_back({ b,false });
		v[b].push_back({ a,false });
	}
	if (fib[k] != n)
		no();
	check(1,k);
	cout << "Yes";
	return 0;
}