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 <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 200000;

int n, co = 0;
int fir[N + 5], nex[N * 2 + 5], vv[N * 2 + 5];
int sz[N + 5];
bool fl[N * 2 + 5] = {0};//�߱�� 
long long ff[N + 5];
int t1, t2;

void dfs1(int o, int fa)
{
	sz[o] = 1;
	for (int i = fir[o]; i; i = nex[i])
	{
		if (vv[i] == fa || fl[i])
		{
			continue;
		}
		dfs1(vv[i], o);
		sz[o] += sz[vv[i]];
	}
}

bool dfs(int o, int fa, int c)
{
	if (sz[o] == ff[c - 1] || sz[o] == ff[c - 2])
	{
		t1 = o;
		t2 = fa;
		for (int i = fir[o]; i; i = nex[i])
		{
			if (vv[i] == fa)
			{
				fl[i % n] = true;
				fl[i % n + n] = true;
				break;
			}
		}
		return true;
	}
	for (int i = fir[o]; i; i = nex[i])
	{
		if (vv[i] == fa || fl[i])
		{
			continue;
		}
		if (dfs(vv[i], o, c))
		{
			return true;
		}
	}
	return false;
}

bool search1(int t, int o)
{
	if (ff[t] <= 3)
	{
		return true;
	}
	dfs1(o, 0);
	if (dfs(o, 0, t))
	{
		int tmp1 = t1, tmp2 = t2;
		if (sz[tmp1] == ff[t - 1])
		{
			return search1(t - 1, tmp1) && search1(t - 2, tmp2);
		}
		else
		{
			return search1(t - 2, tmp1) && search1(t - 1, tmp2);
		}
	}
	else
	{
		return false;
	}
}

int main()
{
	ff[1] = ff[0] = 1;
	scanf("%d", &n);
	for (int i = 2; ; ++i)
	{
		ff[i] = ff[i - 1] + ff[i - 2];
		if (ff[i] >= n)
		{
			co = i;
			break;
		}
	}
	for (int i = 1; i < n; ++i)
	{
		scanf("%d%d", &vv[i + n], &vv[i]);
		nex[i] = fir[vv[i + n]];
		fir[vv[i + n]] = i;
		nex[i + n] = fir[vv[i]];
		fir[vv[i]] = i + n;
	}
	if (n <= 3)
	{
		printf("YES\n");
		return 0;
	}
	if (n != ff[co])
	{
		printf("NO\n");
		return 0;
	}
	bool cc = search1(co, 1);
	if (cc)
	{
		printf("YES\n");
	}
	else
	{
		printf("NO\n");
	}
	return 0;
}