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;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
 
    #ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    #endif

    int n; cin >> n;
    vector<set<int>> G(n);
    for (int i = 0, a, b; i < n - 1; i++) {
        cin >> a >> b;
        a--; b--;
        G[a].insert(b);
        G[b].insert(a);
    }
    unordered_map<int, int> par, sz;
    vector<int> v;

    function<bool(int, int)> check = [&](int i, int n) {
        if (n == 1) return true;
        int x = 1, y = 1;
        while (x + y < n) {
            int tmp = x;
            x = y, y += tmp;
        }
        if (x + y != n) return false;
        v.clear();
        par[i] = -1;

        function<void(int)> dfs = [&](int i) {
            v.push_back(i);
            sz[i] = 1;
            for (int j: G[i]) {
                if (j != par[i]) {
                    par[j] = i;
                    dfs(j);
                    sz[i] += sz[j];
                }
            }
        };

        dfs(i);
        int p = -1, q = -1;
        for (int i: v)  if (sz[i] == x || sz[i] == y)  p = i, q = par[i];
        if (p == -1) return false;
        G[p].erase(q);
        G[q].erase(p);
        if (sz[p] == y) swap(x, y);
        return check(p, x) && check(q, y);
    };

    cout << (check(0, n) ? "YES" : "NO") << endl;

    return 0;
}