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 <algorithm>
#include <random>
#include <chrono>
#include <vector>

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

const int MN = 200005;

int Fibs[55], iFibId[MN];
void Init(int N) {
	Fibs[0] = Fibs[1] = 1;
	for (int i = 2; ; ++i) {
		Fibs[i] = Fibs[i - 1] + Fibs[i - 2];
		if (Fibs[i] > N) break;
		iFibId[Fibs[i]] = i;
	}
}

int N, perm[MN];
std::vector<int> G[MN];

int par[MN], ldf[MN], rdf[MN], dfc;
void DFS(int u, int p) {
	par[u] = p;
	ldf[u] = ++dfc;
	for (int v : G[u]) if (v != p) DFS(v, u);
	rdf[u] = dfc;
}

int bit[MN];
inline void bAdd(int i, int x) { for (; i <= N; i += i & -i) bit[i] += x; }
inline int bQur(int i) { int s = 0; for (; i; i -= i & -i) s += bit[i]; return s; }
inline int Qur(int u) { return bQur(rdf[u]) - bQur(ldf[u] - 1); }

std::vector<int> V[35];
bool Solve(int lay) {
//	printf("Solve(lay = %d, [", lay);
//	for (int x : V[lay]) printf("%d, ", x);
//	puts("])");
	if (lay <= 3) return true;
	int siz = Fibs[lay];
	int csiz1 = Fibs[lay - 1];
	int csiz2 = Fibs[lay - 2];
	for (int i = 0; i < siz; ++i) {
		int u = V[lay][i];
		int sz = Qur(u);
		if (sz == csiz1 || sz == csiz2) {
			int ch1 = (sz == csiz1) ? (lay - 1) : (lay - 2);
			int ch2 = (sz == csiz1) ? (lay - 2) : (lay - 1);
			int lb = ldf[u], rb = rdf[u];
			for (int j = 0; j < siz; ++j) {
				int v = V[lay][j];
				if (lb <= ldf[v] && ldf[v] <= rb)
					V[ch1].push_back(v);
			}
			bool ret = Solve(ch1);
			V[ch1].clear();
			if (ret) {
				bAdd(ldf[par[u]], -sz);
				for (int j = 0; j < siz; ++j) {
					int v = V[lay][j];
					if (ldf[v] < lb || rb < ldf[v])
						V[ch2].push_back(v);
				}
				bool ret2 = Solve(ch2);
				V[ch2].clear();
				bAdd(ldf[par[u]], sz);
				if (ret2) return true;
			}
		}
	}
	return false;
}

int main() {
	Init(200000);
	scanf("%d", &N);
	if (N <= 3) return puts("YES"), 0;
	if (!iFibId[N]) return puts("NO"), 0;
	for (int i = 1; i <= N; ++i) perm[i] = i;
//	std::shuffle(perm + 1, perm + N + 1, rng);
	for (int i = 1; i <= N - 1; ++i) {
		int x, y;
		scanf("%d%d", &x, &y);
		x = perm[x], y = perm[y];
		G[x].push_back(y);
		G[y].push_back(x);
	}
	DFS(1, 0);
	for (int i = 1; i <= N; ++i) bAdd(i, 1);
	int lay = iFibId[N];
	for (int i = 1; i <= N; ++i) V[lay].push_back(i);
	puts(Solve(lay) ? "Yes" : "No");
	return 0;
}