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;

//#define MULTITEST

#define x first
#define y second
#define mp make_pair
#define pb push_back
#define sqr(a) ((a) * (a))
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define forn(i, n) for(int i = 0; i < int(n); i++) 
#define fore(i, l, r) for(int i = int(l); i < int(r); i++)

typedef long long li;
typedef long double ld;
typedef pair<int, int> pt;

template <class A, class B> ostream& operator << (ostream& out, const pair<A, B> &a) {
	return out << "(" << a.x << ", " << a.y << ")";
}

template <class A> ostream& operator << (ostream& out, const vector<A> &v) {
	out << "[";
	forn(i, sz(v)) {
		if(i) out << ", ";
		out << v[i];
	}
	return out << "]";
}

mt19937 rnd(time(NULL));

const int INF = int(1e9);
const li INF64 = li(1e18);
const int MOD = int(1e9) + 7;
const ld EPS = 1e-9;
const ld PI = acos(-1.0);

const int N = 200 * 1000 + 13;

int n;
vector<int> g[N];

bool read () {
	if (scanf("%d", &n) != 1)
		return false;
	forn(i, n)
		g[i].clear();
	forn(i, n - 1){
		int v, u;
		scanf("%d%d", &v, &u);
		--v, --u;
		g[v].pb(u);
		g[u].pb(v);
	}
	return true;
}

int fib[N];
int fibnum[N];
int siz[N];

void init(int v, int p){
	siz[v] = 1;
	for (int u : g[v]) if (u != p){
		init(u, v);
		siz[v] += siz[u];
	}
}

int dfs(int v, int p, int siz1, int siz2){
	for (int u : g[v]) if (u != p){
		if (siz[u] == siz1 || siz[u] == siz2){
			g[v].erase(find(all(g[v]), u));
			g[u].erase(find(all(g[u]), v));
			return u;
		}
		int res = dfs(u, v, siz1, siz2);
		if (res != -1) return res;
	}
	return -1;
}

bool calc(int v){
	init(v, -1);
	int cnt = siz[v];
	if (fibnum[cnt] == -1) return false;
	if (cnt <= 2) return true;
	int nrt = dfs(v, -1, fib[fibnum[cnt] - 1], fib[fibnum[cnt] - 2]);
	if (nrt == -1) return false;
	return calc(v) && calc(nrt);
}

void solve() {
	forn(i, n) if (sz(g[i]) > 30){
		puts("NO");
		return;
	}
	fib[0] = 1;
	fib[1] = 2;
	fore(i, 2, 26) fib[i] = fib[i - 1] + fib[i - 2];
	memset(fibnum, -1, sizeof(fibnum));
	forn(i, 26) fibnum[fib[i]] = i;
	puts(calc(0) ? "YES" : "NO");
}

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
	
	int tt = clock();
	
#endif
	
	cerr.precision(15);
	cout.precision(15);
	cerr << fixed;
	cout << fixed;

#ifdef MULTITEST
	int tc;
	scanf("%d", &tc);
	while(tc--){
		read();
#else
	while(read()) {	
#endif
		solve();
		
#ifdef _DEBUG
	cerr << "TIME = " << clock() - tt << endl;
	tt = clock();
#endif

	}
}