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;
// curiosity
const int N = 2e5 + 5;
int n, siz[N];
vector<int> fib;
vector<pair<int, bool> > G[N];
void NO() { cout << "NO\n"; exit(0); }
void GetSize(int u, int fa)
{
    siz[u] = 1;
    for(pair<int, bool> e : G[u])
    {
        if(e.second) continue;
        int v = e.first;
        if(v == fa) continue;
        GetSize(v, u);
        siz[u] += siz[v];
    }
}
void CutEdge(int u, int fa, int k, int &pu, int &pv, int &kd)
{
    for(pair<int, bool> e : G[u])
    {
        if(pu) return;
        if(e.second) continue;
        int v = e.first;
        if(v == fa) continue;
        if(siz[v] == fib[k - 1] || siz[v] == fib[k - 2])
        {
            pu = u; pv = v;
            kd = (siz[v] == fib[k - 1]) ? k - 1 : k - 2;
            break;
        }
        CutEdge(v, u, k, pu, pv, kd);
    }
}
void Check(int u, int k)
{
//    cout << "Check " << u << ' ' << k << endl;
    if(k <= 1) return;
    GetSize(u, 0);
    int pu = 0, pv = 0, kd = 0;
    CutEdge(u, 0, k, pu, pv, kd);
//    cout << pu << ' ' << pv << ' ' << kd << endl;
    if(!pu) NO();
    for(pair<int, bool> &e : G[pu])
        if(e.first == pv) e.second = true;
    for(pair<int, bool> &e : G[pv])
        if(e.first == pu) e.second = true;
    Check(pv, kd);
    Check(pu, (kd == k - 1) ? k - 2 : k - 1);
}
int main()
{
    ios::sync_with_stdio(false);
    cin >> n;
    fib.push_back(1);
    fib.push_back(1);
    for(int i = 1; ; i++)
    {
        if(fib[i] >= n) break;
        int new_fib = fib[i] + fib[i - 1];
        fib.push_back(new_fib);
    }
    for(int i = 1; i < n; i++)
    {
        int u, v;
        cin >> u >> v;
        G[u].push_back(make_pair(v, false));
        G[v].push_back(make_pair(u, false));
    }
    if(fib[fib.size() - 1] != n) NO();
    Check(1, fib.size() - 1);
    cout << "YES\n";
    return 0;
}