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;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair<int,int> pi;
typedef pair<ll,ll> pl;
#define F first
#define S second
#define endl '\n'
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()
#ifdef dremix
    #define p(x) cerr<<#x<<" = "<<x<<endl;
    #define p2(x,y) cerr<<#x<<" , "<<#y<<" = "<<x<<" , "<<y<<endl;
    #define pp(x) cerr<<#x<<" = ("<<x.F<<" , "<<x.S<<")"<<endl;
    #define pv(x) cerr<<#x<<" = {";for(auto u : x)cerr<<u<<", ";cerr<<"}"<<endl;
    #define ppv(x) cerr<<#x<<" = {";for(auto u : x)cerr<<u.F<<"-"<<u.S<<", ";cerr<<"}"<<endl;
#else
    #define p(x)
    #define p2(x,y)
    #define pp(x)
    #define pv(x)
    #define ppv(x)
#endif
#define fastio ios_base::sync_with_stdio(false);cin.tie(nullptr);
const int maxp = 22;
const ld EPS = 1e-18;
const ll INF = 2e18;
const int MOD = 1e9+7;
const int N = 2e5+1;
void yesno(bool a){if(a){cout<<"YES"<<endl;}else{cout<<"NO"<<endl;}}
ll power(ll x, ll n, ll m){if(n==0)return 1;ll u=power(x,n/2,m);u=(u*u)%m;if(n%2)u=(u*x)%m;return u;}
/// * int overflow, array bounds
/// * special cases (n=1?)
/// * do something - write observations down
/// * change approaches

vector<set<int> > a(N);
int dp[27],cnt[N],par[N];
vector<pi> arr;

void dfs(int s, int e){
    cnt[s] = 1;
    par[s] = e;
    for(auto x : a[s])
    if(x != e){
        dfs(x,s);
        cnt[s] += cnt[x];
    }
    arr.push_back({cnt[s],s});
}

bool solve(int s, int idx){
    if(idx <= 3)
        return true;

    int p1 = dp[idx-1],p2 = dp[idx-2];
    arr.clear();
    dfs(s,s);

    assert(cnt[s] == dp[idx]);

    p2(p1,p2)
    for(auto x : arr){
        if(x.F == p1){
            a[x.S].erase(par[x.S]);
            a[par[x.S]].erase(x.S);
            int yy = par[x.S];
            return solve(x.S,idx-1)&solve(yy,idx-2);
        }
        if(x.F == p2){
            a[x.S].erase(par[x.S]);
            a[par[x.S]].erase(x.S);
            int yy = par[x.S];
            return solve(x.S,idx-2)&solve(yy,idx-1);
        }
    }
    return false;
}

int main (){
fastio;

int m = 27;

dp[0] = dp[1] = 1;

int i;

for(i=2;i<m;i++){
    dp[i] = dp[i-1] + dp[i-2];
}

for(i=0;i<m;i++){
    p2(i,dp[i])
}

int n;
cin>>n;

int pos = 1;
bool ok = 0;

while(pos<m && dp[pos] < n){
    pos++;
}

if(pos<m && dp[pos] == n)
    ok = 1;

if(!ok){
    yesno(0);
    return 0;
}

/// solve recursively?

for(i=0;i<n-1;i++){
    int x,y;
    cin>>x>>y;
    a[x].insert(y);
    a[y].insert(x);
}

yesno(solve(1,pos));

/*

13
1 9
9 13
9 6
13 5
6 7
1 2
7 3
9 8
8 10
13 12
5 4
12 11

*/


}