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 int long long

using namespace std;

const int maxn = 200005;
set<int> adj[maxn];
int sz[maxn];
set<pair<int,int>> s;

void calcSizes(int u, int par) {
    sz[u] = 1;
    for (int v : adj[u]) {
        if (v == par) continue;
        calcSizes(v, u);
        sz[u] += sz[v];
    }
}

pair<int,int> findEdge(int u, int par, int nodes) {
    pair<int,int> e = {-1,-1};
    for (int v : adj[u]) {
        if (v == par) continue;
        pair<int,int> p = findEdge(v,u,nodes);
        if (p.first != -1) e = p;
        if (s.count({nodes-sz[v],sz[v]}) != 0 || s.count({sz[v],nodes-sz[v]}) != 0) e = {u,v};
    }
    return e;
}

void rec(int r) {
    calcSizes(r,r);
    if (sz[r] == 1) return;
    pair<int,int> e = findEdge(r,r,sz[r]);
    if (e.first == -1) {
        cout << "NO\n";
        exit(0);
    }
    adj[e.first].erase(e.second);
    adj[e.second].erase(e.first);
    rec(e.first);
    rec(e.second);
}

signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int x = 1;
    int y = 1;
    s.insert({x,y});
    for (int i = 0; i < 30; i++) {
        int tmp = y;
        y = x+y;
        x = tmp;
        s.insert({x,y});
    }
    int n;
    cin >> n;
    for (int i = 0; i < n-1; i++) {
        int u, v;
        cin >> u >> v;
        adj[u].insert(v);
        adj[v].insert(u);
    }
    rec(1);
    cout << "YES\n";
}
/*

*/