#include<bits/stdc++.h>
#include<atcoder/all>
using namespace std;
using namespace atcoder;
#define rep(i,n)for (int i = 0; i < int(n); ++i)
#define rrep(i,n)for (int i = int(n)-1; i >= 0; --i)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
template<class T> void chmax(T& a, const T& b) {a = max(a, b);}
template<class T> void chmin(T& a, const T& b) {a = min(a, b);}
using ll = long long;
using P = pair<int,int>;
using VI = vector<int>;
using VVI = vector<VI>;
using VL = vector<ll>;
using VVL = vector<VL>;
using mint = modint1000000007;

int n;
int c[11];
int solve() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int cs = 0;
    rep(i, 10) cs += c[i] > 0;
    if (cs == 1) {
        rep(i, 10) if (c[i]) {
            return ((mint(10).pow(n) - 1) / 9 * i).val();
        }
        return 0;
    }
    if (n <= 3) {
        ll g = 0;
        string s;
        rep(i, 10) s.append(c[i], '0' + i);
        do {
            g = gcd(g, stoll(s));
        } while(next_permutation(all(s)));
        return g;
    }

    if (c[3] + c[6] + c[9] == n) {
        int r = (c[3] + 2LL * c[6] + 3LL * c[9]) % 9;
        if (r % 9 == 0) return 27;
        else if (r % 3 == 0) return 9;
        else return 3;
    }
    int r9 = 0;
    rep(i, 10) r9 = (r9 + (ll)i * c[i]) % 9;
    bool b2 = true, b4 = true;
    rep(i, 10) {
        if (i % 2 && c[i]) b2 = false;
        if (i % 4 && c[i]) b4 = false;
    }
    int g = 1;
    if (r9 % 3 == 0) g *= 3;
    if (r9 % 9 == 0) g *= 3;
    if (b2) g *= 2;
    if (b4) g *= 2;
    return g;
}

int main() {
    cin >> n;
    rep(i, 9) cin >> c[i + 1];
    cout << solve() << endl;
}