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;

#define FAST ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define mp make_pair
#define pb push_back
#define lp(i,s,f) for(ll i = s; i < ll(f); i++)
#define inF freopen("input.in", "r", stdin);
#define outF freopen("output.in", "w", stdout);
#define endl '\n'
#define MOD 1000000007
#define mm(arr) memset(arr, 0, sizeof(arr))
#define F first
#define S second

const int N = 2e5 + 5;
vector<int> adj[N];
int parent[N];
int sz[N];
bool visited[N];
int n;
vector<int> comp;
bool fib[N];
void dfs(int s, int p){
    parent[s] = p;
    comp.pb(s);
    visited[s] = 1;
    for(int i = 0; i < adj[s].size(); i++){
        if(adj[s][i] == p)continue;
        dfs(adj[s][i], s);
        sz[s] += sz[adj[s][i]];
    }
}
void cut(int a, int b){
    for(int i = 0; i < adj[a].size(); i++){
        if(adj[a][i] == b){
            adj[a].erase(adj[a].begin() + i);
            break;
        }
    }
    swap(a, b);
    for(int i = 0; i < adj[a].size(); i++){
        if(adj[a][i] == b){
            adj[a].erase(adj[a].begin() + i);
            break;
        }
    }
}
bool getEdge(){
    for(int i = 0; i < comp.size(); i++){
        if(fib[sz[comp[i]]] && fib[(int)comp.size() - sz[comp[i]]]){
            cut(comp[i], parent[comp[i]]);
            return 1;
        }
    }
    return 0;
}
bool check(){
    for(int i = 0; i < n; i++){
        visited[i] = 0;
        parent[i] = 0;
        sz[i] = 1;
    }
    bool ok = 1;
    for(int i = 0; i < n; i++){
        if(visited[i])continue;
        comp.clear();
        dfs(i, -1);
        if(comp.size() > 3){
            if(!fib[(int)comp.size()]){
                return 0;
            }
            if(!getEdge()){
                return 0;
            }
            ok = 0;
        }
    }
    if(!ok){
        return check();
    }
    return 1;
}
int main(){
    //FAST
    cin >> n;
    fib[1] = 1;
    int a = 1, b = 1;
    while(true){
        if(a + b > n)break;
        fib[a + b] = 1;
        int tmp = a;
        a = a + b;
        b = tmp;
    }
    for(int i = 0; i < n - 1; i++){
        int a, b;
        cin >> a >> b;
        a--; b--;
        adj[a].pb(b);
        adj[b].pb(a);
    }
    cout << (check()? "YES": "NO");
    return 0;
}