#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

map<ll,ll> mp; //prime number --> the occurence of this prime

ll a[106];

void f1(int num,int cnt) {
    for (int i=2;i<=num;++i) {
        while (num % i == 0) {
            mp[i] += cnt;
            num /= i;
        }
    }
}

bool is_p[1006];

int main () {
    for (int i=2;i<1006;++i) {
        if (!is_p[i]) {
            for (int j=i+i;j<1006;j+=i) is_p[j] = true;
        }
    }
    //is_p[i] == false means i is a prime
    int n;
    cin >> n;
    for (int i=0;i<n;++i) {
        cin >> a[i];
    }
    for (int times=0;times<100;++times) {
        for (int i=0;i<n;++i) {
            for (int j=i+1;j<n;++j) {
                ll gcd = __gcd(a[i],a[j]);
                if (gcd != 1) {
                    int cnt=0;
                    for (int k=0;k<n;++k) {
                        if (a[k]%gcd == 0) {
                            cnt++;
                            a[k] /= gcd;
                        }
                    }
                    f1(gcd,cnt);
                }
            }
        }
    }
    //after the process above, numbers left must be one of the following forms: p, p^2, p^3, p^4, p*q, and every prime factor in a[i] are distinct!!!
    for (int i=0;i<n;++i) {
        if (a[i] != 1) {
            for (int x=2;x<=40;++x) {
                if (x*x*x*x == a[i]) {
                    a[i] = 1;
                    mp[x] += 4;
                    break;
                }
                if (x*x*x == a[i]) {
                    a[i] = 1;
                    mp[x] += 3;
                    break;
                }
                if (x*x == a[i]) {
                    a[i] = 1;
                    mp[x] += 2;
                    break;
                }
            }
            if (a[i] != 1) {
                if (!is_p[ a[i] ]) mp[ a[i] ]+=1;
                else {
                    //since I do not know what exactly p & q is, so I assume that one is a[i], the other is -a[i]
                    mp[ a[i] ]  = 1;
                    mp[ -a[i] ] = 1;
                }
            }
        }
    }
    ll ans=1;
    for (auto i:mp) {
        ans *= (i.second+1);
        ans %= 880301;
    }
    cout << ans << endl;
}

#include <cstdio>
int main() {
    printf("3\n4 8 16\n");
}