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;

vector<vector<array<int, 2>>> g(200200);
vector<int> f(30, 1), sz(200200), en(200200, 1);
int a, b, c;

void dfs(int i, int d, int id, int ff){
 sz[i] = 1;
 for(auto [j, ij] : g[i]) if(j != d && en[ij])
  dfs(j, i, ij, ff), sz[i] += sz[j];
 if(sz[i] == f[ff-1] || sz[i] == f[ff-2]) a = i, b = d, c = id;
}

int ses(int i, int ff){
 a = b = c = 0;
 dfs(i, 0, 0, ff);
 if(sz[i] == 1) return 1;
 if(sz[i] != f[ff] || !c) return 0;
 en[c] = 0;
 int aa = a, bb = b;
 if(sz[aa] != f[ff-1]) return ses(bb, ff-1) && ses(aa, ff-2);
 if(sz[aa] == f[ff-1]) return ses(aa, ff-1) && ses(bb, ff-2);
}

int main(){
 for(int i = 2; i < 30; ++i) f[i] = f[i-1] + f[i-2];
 int n; cin >> n;
 for(int i = 1; i < n; ++i){
  int a, b; cin >> a >> b;
  g[a].push_back({b, i});
  g[b].push_back({a, i});
 }
 int ff = 0;
 while(f[ff] < n) ++ff;
 puts((f[ff] == n && ses(1, ff))? "yes" : "no");
}