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:
#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
#include <chrono>
#include <random>
#include <unordered_set>

using namespace std;

typedef long long ll;

void solve() {
    int n;
    cin >> n;
    vector<vector<int>> g(n);
    for (int i = 0; i < n - 1; i++) {
        int f, t;
        cin >> f >> t;
        f--;
        t--;
        g[f].push_back(t);
        g[t].push_back(f);
    }

    vector<int> fib(30);
    fib[0] = fib[1] = 1;
    for (int i = 2; i < 30; i++) {
        fib[i] = fib[i - 1] + fib[i - 2];
    }

    vector<int> cnt(n);

    vector<int> fob(n);

    function<bool(int, int)> go = [&](int u, int h) {
        if (h == 0 || h == 1) {
            bool ok = 1;
            for (int to : g[u]) {
                if (fob[to] == 0)
                    ok = 0;
            }
            return ok;
        }

        function<void(int, int)> dfs1 = [&](int v, int pred) {
            cnt[v] = 1;
            for (int to : g[v]) {
                if (to ^ pred && fob[to] != 1) {
                    dfs1(to, v);
                    cnt[v] += cnt[to];
                }
            }
        };

        dfs1(u, u);
        bool ok = 0;
        function<void(int, int)> dfs2 = [&](int v, int pred) {
            for (int to : g[v]) {
                if (to ^ pred && fob[to] != 1) {
                    int f = cnt[u] - cnt[to];
                    int s = cnt[to];
                    bool sw = 0;
                    if (f > s) {
                        swap(f, s);
                        sw = 1;
                    }
                    if (fib[h - 1] == s && fib[h - 2] == f) {
                        fob[v] = 1;
                        bool ui = go(to, (sw ? h - 2 : h - 1));
                        fob[v] = 0;
                        fob[to] = 1;
                        bool vi = go(v, (sw ? h - 1 : h - 2));
                        fob[to] = 0;
                        ok = (ui & vi);
                    }
                }
            }

            if (!ok) {
                for (int to : g[v]) {
                    if (to ^ pred && fob[to] != 1) {
                        dfs2(to, v);
                    }
                }
            }
        };

        dfs2(u, u);
        return ok;
    };

    int pos = lower_bound(fib.begin(), fib.end(), n) - fib.begin();
    if (fib[pos] != n) {
        cout << "No";
        return;
    }
    cout << (go(0, pos) ? "Yes" : "No");
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
#if defined(_DEBUG)
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    int q = 1;
    //cin >> q;
    for (; q > 0; q--) {
        solve();
        //cout << endl;
    }
}