#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;

#define pb push_back

const ld pi = 3.14159265358979323846;
// const int mod = 998244353;
const ll INF = 1e18;

template<typename T>
T chmax(T a, T b) {
    return a > b ? a : b;
}

template<typename T>
T chmin(T a, T b) {
    return a > b ? b : a;
}

#define L(x) (1 << (x))

const double PI = acos(-1.0);
const int Maxn = 13301500;

double ax[Maxn], ay[Maxn], bx[Maxn], by[Maxn];
char sa[Maxn/2],sb[Maxn/2];
int sum[Maxn];
int x1[Maxn],x2[Maxn];

int revv(int x, int bits)
{
    int ret = 0;
    for (int i = 0; i < bits; i++)
    {
        ret <<= 1;
        ret |= x & 1;
        x >>= 1;
    }
    return ret;
}

void fft(double * a, double * b, int n, bool rev)
{
    int bits = 0;
    while (1 << bits < n) ++bits;
    for (int i = 0; i < n; i++)
    {
        int j = revv(i, bits);
        if (i < j)
            swap(a[i], a[j]), swap(b[i], b[j]);
    }
    for (int len = 2; len <= n; len <<= 1)
    {
        int half = len >> 1;
        double wmx = cos(2 * PI / len), wmy = sin(2 * PI / len);
        if (rev) wmy = -wmy;
        for (int i = 0; i < n; i += len)
        {
            double wx = 1, wy = 0;
            for (int j = 0; j < half; j++)
            {
                double cx = a[i + j], cy = b[i + j];
                double dx = a[i + j + half], dy = b[i + j + half];
                double ex = dx * wx - dy * wy, ey = dx * wy + dy * wx;
                a[i + j] = cx + ex, b[i + j] = cy + ey;
                a[i + j + half] = cx - ex, b[i + j + half] = cy - ey;
                double wnx = wx * wmx - wy * wmy, wny = wx * wmy + wy * wmx;
                wx = wnx, wy = wny;
            }
        }
    }
    if (rev)
    {
        for (int i = 0; i < n; i++)
            a[i] /= n, b[i] /= n;
    }
}

int solve(int a[],int na,int b[],int nb,int ans[])
{
    int len = max(na, nb), ln;
    for(ln=0; L(ln)<len; ++ln);
    len=L(++ln);
    for (int i = 0; i < len ; ++i)
    {
        if (i >= na) ax[i] = 0, ay[i] =0;
        else ax[i] = a[i], ay[i] = 0;
    }
    fft(ax, ay, len, 0);
    for (int i = 0; i < len; ++i)
    {
        if (i >= nb) bx[i] = 0, by[i] = 0;
        else bx[i] = b[i], by[i] = 0;
    }
    fft(bx, by, len, 0);
    for (int i = 0; i < len; ++i)
    {
        double cx = ax[i] * bx[i] - ay[i] * by[i];
        double cy = ax[i] * by[i] + ay[i] * bx[i];
        ax[i] = cx, ay[i] = cy;
    }
    fft(ax, ay, len, 1);
    for (int i = 0; i < len; ++i)
        ans[i] = (int)(ax[i] + 0.5);
    return len;
}

string mul(string sa,string sb)
{
    int l1,l2,l;
    int i;
    string ans;
    memset(sum, 0, sizeof(sum));
    l1 = sa.size();
    l2 = sb.size();
    for(i = 0; i < l1; i++)
        x1[i] = sa[l1 - i - 1]-'0';
    for(i = 0; i < l2; i++)
        x2[i] = sb[l2-i-1]-'0';
    l = solve(x1, l1, x2, l2, sum);
    for(i = 0; i<l || sum[i] >= 10; i++) // 进位
    {
        sum[i + 1] += sum[i] / 10;
        sum[i] %= 10;
    }
    l = i;
    while(sum[l] <= 0 && l> 0)    l--; // 检索最高位
    for(i = l; i >= 0; i--)    ans+=sum[i] + '0'; // 倒序输出
    return ans;
}



/*
作者: lswtn
链接: https://lswtn.github.io/2024/04/27/FFT_high_precision_mul/index.html
来源: lswtn
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
*/

bool cmp(string a, string b){
    if(a.size() > b.size()) return 0;
    if(a.size() < b.size()) return 1;

    for(int i = 0; i < a.size(); i++){
        if(a[i] > b[i]) return 0;
        else if(b[i] > a[i]) return 1;
    }
    return 1;
}

void solve(){
    int n;
    cin >> n;

    vector<string> a;
    for(int i = 0; i < n; i++){
        string s;
        cin >> s;

        a.pb(s);
    }

    sort(a.begin(), a.end(), cmp);

    while(1){
        vector<string> b;

        for(int i = 1; i < a.size(); i++){
            if(a[i] == a[i - 1]){
                b.pb(mul(a[i], a[i - 1]));
                i++;
            }
            else{
                b.pb(a[i - 1]);
            }
        }
        if(a.size() > 1 && a.back() != a[a.size() - 2]){
            b.pb(a.back());
        }

        sort(b.begin(), b.end(), cmp);

        if(b.size() == a.size() || a.size() == 1){
            a = b;
            break;
        }
        a = b;
    }

    cout << a.back() << endl;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    int t = 1;
    // cin >> t;

    while(t--){
        solve();
    }

    return 0;
}