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fibonacci_gcd.cpp
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/*
*
* J1K7_7
*
*/
//Ad Infinitum 9 - Math Programming Contest
//https://www.hackerrank.com/contests/infinitum9/challenges/fibonacci-gcd
#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <set>
#include <cstring>
#include <list>
#include <map>
#include <iomanip>
#include <algorithm>
#include <functional>
#include <utility>
#include <bitset>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstdio>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<long long> vll;
#define l(x) (x << 1) + 1
#define r(x) (x << 1) + 2
#define mid(l, r) ((l + r) >> 1)
#define mp make_pair
#define pb push_back
#define all(a) a.begin(),a.end()
#define pr(n) printf("%d",n)
#define s(n) scanf("%d",&n)
#define debug(x) {cerr <<#x<<" = " <<x<<"\n"; }
#define debug2(x, y) {cerr <<#x<<" = " <<x<<", "<<#y <<" = " <<y <<"\n";}
#define ss second
#define ff first
#define m0(x) memset(x,0,sizeof(x))
#define snuke(c,itr) for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)
const int mod=1e9+7;
const ll mx_ll = numeric_limits<ll> :: max();
const int mx_int = numeric_limits<int> :: max();
const long double PI = (long double)(3.1415926535897932384626433832795);
inline bool ispow2(int x){return (x!=0 && (x&(x-1))==0);} //0 or 1
int msb(unsigned x){ union { double a; int b[2]; }; a = x; return (b[1] >> 20) - 1023; }
template<class T>
inline void cinarr(T a, int n){ for (int i=0;i<n;++i) cin >> a[i];}
inline ll powmod(ll a,ll b) {ll res = 1; while(b){if(b&1) res = (res*a)%mod;a = (a*a)%mod;b >>= 1;}return res;}
inline ll gcd(ll a,ll b){ll t;while(b){a=a%b;t=a;a=b;b=t;}return a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
inline int nextint(){ int x; scanf("%d",&x); return x; }
inline ll nextll(){ ll x; scanf("%lld",&x); return x; }
map <ll ,ll > F;
ll f(ll n)
{
F[0] = F[1] = 1;
if (F.count(n))
return F[n];
long k=n/2;
if (n%2==0)
{
return F[n] = (f(k)*f(k) + f(k-1)*f(k-1)) % mod;
}
else
{
return F[n] = (f(k)*f(k+1) + f(k-1)*f(k)) % mod;
}
}
// query nth fibonacci number in logn * log(logn)
// (n == 0 ) ? 0 : f(n-1)
int main()
{
ios_base::sync_with_stdio(false); cin.tie(0);
int n = nextint();
ll g = nextll();;
for(int i = 1; i < n; i++)
{
ll x = nextll();
g = gcd(g , x);
}
printf ( "%lld\n", f(g-1) );
return 0;
}