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:
﻿//#pragma optimization_level 3

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <chrono>
#include <ctime>
#include <queue>
#include <math.h>
#include <deque>
#include <stack>
#include <iomanip>
#include <assert.h>
#include <stdio.h>
#include <cstring>
#include <random>

using namespace std;

#define ll long long
#define ull unsigned long long
#define ld long double
#define pii pair<int, int>
#define pld pair<ld, ld>
#define ti tuple<int, int, int>
#define vi vector <int>
#define vpi vector <pii>
#define vld vector <ld>
#define try tr
#define left left11
#define right right11
#define lb lower_bound
#define ub upper_bound
#define sz(c) (int)(c).size()
#define all(c) (c).begin(), (c).end()
#define pb push_back
#define pf push_front
#define sqr(a) ((a) * (a))
#define dbg(x) cout << #x << " : " << x << endl;
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

#define x first
#define y second

template<typename T1, typename T2>
inline bool amin(T1 &a, T2 b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

template<typename T1, typename T2>
inline bool amax(T1 &a, T2 b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

template<typename T1, typename T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p) {
    os << p.first << ' ' << p.second;
    return os;
}

template<typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) {
    is >> p.first >> p.second;
    return is;
}

template<typename T>
istream &operator>>(istream &is, vector<T> &v) {
    for (auto &u : v) is >> u;
    return is;
}

template<typename T>
ostream &operator<<(ostream &os, vector<T> &v) {
    for (auto &u : v) os << u << ' ';
    return os;
}

const int N = 2e5, inf = 1e9, pi = 30;

vpi v[N];

int sz[N], f[100], ban[N];

map <int, int> res[N];

int calc(int x, int p) {
    sz[x] = 1;
    for (auto u : v[x]) {
        if(u.x == p || ban[u.y])
            continue;
        sz[x] += calc(u.x, x);
    }
    return sz[x];
}

bool check(int x, int p, int cur) {
    if(cur <= 1) {
        return (sz[x] == 1);
    }
    vpi e;
    for (auto u : v[x]) {
        if(u.x == p || ban[u.y])
            continue;
        if(check(u.x, x, cur))
            return true;
        if(sz[u.x] == f[cur - 1] || sz[u.x] == f[cur - 2])
            e.pb(u);
    }

    for (auto u : e) {

        ban[u.y] = 1;

        calc(x, -1);
        calc(u.x, -1);

        if(sz[u.x] == f[cur - 1] && check(u.x, -1, cur - 1) && check(x, -1, cur - 2)) {
            return true;
        }
        if(sz[u.x] == f[cur - 2] && check(u.x, -1, cur - 2) && check(x, -1, cur - 1))
            return true;

        ban[u.y] = 0;
    }
    return false;

}

void solve() {
    f[0] = f[1] = 1;
    for (int i = 2; i < pi; i++)
        f[i] = f[i - 1] + f[i - 2];
    int n;
    cin >> n;
    for (int i = 1; i < n; i++) {
        int a, b;
        cin >> a >> b;
        a--, b--;
        v[a].pb({b, i});
        v[b].pb({a, i});
    }

    int id = -1;
    for (int i = 0; i < pi; i++) {
        if(f[i] == n) {
            id = i;
        }
    }

    if(id == -1) {
        cout << "NO";
        return;
    }

    calc(0, -1);

    if(check(0, -1, id))
        cout << "YES";
    else
        cout << "NO";

}

signed main() {
#ifdef _LOCAL
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    fast
    solve();
}