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 long long LL;

const int MN = 4e5 + 5;

struct Edge {
	int to , nxt;
} E[MN << 1];

int N , f[MN] , tot = 1 , size[MN] , v0[MN] , vis[MN] , head[MN];

inline void AddEdge(int u , int v) {
	E[++tot] = (Edge) {v , head[u]};
	head[u] = tot;
}
inline int dfs(int x, int res1, int res2) {
	int ret = -1;
	size[x] = 1, v0[x] = 1;
	for (int i = head[x]; i; i = E[i].nxt)
		if (!vis[i] && !v0[E[i].to]) {
			int tmp = dfs(E[i].to , res1, res2);
			if (ret == -1) ret = tmp;
			size[x] += size[E[i].to];
			if (size[E[i].to] == res1)
				ret = i;
			if (size[E[i].to] == res2)
				ret = i ^ 1;
		}
	v0[x] = 0;
	return ret;
}
inline bool solve(int x , int now) {
	if (now <= 1) return true;
	int u = dfs(x , f[now - 2] , f[now - 1]);
	if (u == -1) return false;
	vis[u] = vis[u ^ 1] = 1;
	return solve(E[u].to , now - 2) && solve(E[u ^ 1].to , now - 1);
}

int main() {
	
	scanf("%d" , &N);
	if (N == 1) {
		printf("YES\n");
		return 0;
	}
	for (int i = 1 , u , v; i < N; ++i) {
		scanf("%d%d" , &u , &v);
		AddEdge(u , v) , AddEdge(v , u);		
	}
	f[0] = f[1] = 1;
	for (int i = 2; i < 30; ++i) {
		f[i] = f[i - 1] + f[i - 2];
		if (f[i] > N) {
			printf("NO\n");
			break;
		}
		if (f[i] == N) {
			if (solve(1 , i)) printf("YES\n");
			else printf("NO\n");
			return 0;
		}
	} 
	return 0;
}