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>
#define long long long
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int N=1e6,inf=1e9,mod=1e9+7;

set<int>ad[N+2];
int f[N+2],ln[N+2],fib[N+2],vs[N+2],sub[N+2],pt,an[N+2];

void no()
{
  cout<<"NO"<<endl,exit(0);
}
void reckon(int nd,int pr)
{
  sub[nd]=1;ln[++pt]=nd;an[nd]=pr;
  for(auto x:ad[nd])
  {
    if(x==pr)continue;
    reckon(x,nd);sub[nd]+=sub[x];
  }
}
int a,b;
void dfs(int rt,int nd,int pr)
{
  if(fib[sub[nd]]&&fib[sub[rt]-sub[nd]])a=nd,b=pr;
  if(an[nd]!=pr)assert(0);
  for(auto x:ad[nd])
  {
    if(x==pr)continue;
    dfs(rt,x,nd);
  }
}
bool centroid(int nd)
{
  pt=0;reckon(nd,-1);
  if(fib[sub[nd]]==0)no();
  if(sub[nd]<=3)return true;

  a=-1;
  dfs(nd,nd,-1);
  if(a==-1)no();int b=an[a];
  ad[a].erase(b);ad[b].erase(a);
  return (centroid(a)&centroid(b));
}

int main()
{
  ios_base::sync_with_stdio(0);cin.tie(0);

  f[0]=1;
  f[1]=1;fib[1]=1;
  for(int i=2;i<=26;i++)
  {
    f[i]=f[i-1]+f[i-2];
    fib[f[i]]=1;
  }
  //cout<<f[26]<<endl;

  int n;cin>>n;
  if(fib[n]==0)cout<<"NO"<<endl,exit(0);
  for(int i=1;i<n;i++)
  {
    int u,v;cin>>u>>v;
    ad[u].insert(v);ad[v].insert(u);
  }
  if(centroid(1))cout<<"YES"<<endl;
  else cout<<"NO"<<endl;

  
  return 0;
}