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:
/*
#pragma GCC optimize("O2")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2,fma")
*/
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int , int> pii;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int maxn = 2e5 + 100;
const ll mod = 1e9+7;
const ld PI = acos((ld)-1);

#define pb push_back
#define endl '\n'
#define dokme(x) cout << x , exit(0)
#define migmig ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define ms(x , y) memset(x , y , sizeof x)
ll pw(ll a, ll b, ll md = mod){ll res = 1;while(b){if(b&1){res=(a*res)%md;}a=(a*a)%md;b>>=1;}return(res);}

int n;
vector < pii > adj[maxn];
int sz[maxn];
bool hide[maxn];
bool fib[maxn];
int bg[maxn];
int par[maxn];
int big , target , tar;

void plant(int v){
	sz[v] = 1;
	for(auto [u , i] : adj[v]){
		if(!hide[i] and u ^ par[v])
			par[u] = v , plant(u) , sz[v] += sz[u];
	}
	if(sz[v] == target or sz[v] == tar)big = v;
}

int fnd(int v , int p){
	return(lower_bound(adj[v].begin() , adj[v].end() , pii(p , 0))->second);
}

void solve(int v, int n){
	if(!fib[n])dokme("NO");
	if(n <= 3)return;
	par[v] = 0;
	target = bg[n];
	tar = n - bg[n];
	big = 0;
	plant(v);
	v = big;
	if(!v)dokme("NO");
	hide[fnd(v , par[v])] = 1;
	int siz = sz[v];
	solve(par[v] , n-sz[v]);
	solve(v , siz);
}

int32_t main(){
	migmig;
	int a = 1 , b = 1;
	while(b < maxn){
		fib[a] = fib[b] = 1;
		b = a + b;
		a = b - a;
		if(b < maxn)bg[b] = a;
	}
	cin >> n;
	for(int i = 1 ; i < n ; i ++){
		int u , v;
		cin >> u >> v;
		adj[u].pb({v , i});
		adj[v].pb({u , i});
	}
	for(int i = 1 ; i <= n ; i ++)
		sort(adj[i].begin() , adj[i].end());
	solve(1 , n);
	cout << "YES";
	return(0);
}