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为了求解问题，我们首先分析每个数a_i的质因数分解情况，然后将它们相乘得到sum，最后计算sum的约数个数。具体步骤如下：

通过上述步骤，可以高效地计算出sum的约数个数，并对结果取模。

代码实现：

#include 
   
    using namespace std;const ll mod = 998244353;const int maxn = 1e4 + 5;const double eps = 1e-6;int n;int cnt = 0, cur = 0;ll ans = 1;struct node {    ll x, num;} a[MAXN];struct prime {    ll x, num;} p[MAXN];ll four(ll x) {    ll y = pow(x, 1.0 / 4) + eps;    if (y * y * y * y == x) {        bool flag = 0;        for (int i = 1; i <= cnt; i++) {            if (y == p[i].x) {                flag = 1;                p[i].num += 4;                break;            }        }        if (!flag) {            p[++cnt].x = y;            p[cnt].num = 4;        }        return 1;    }    return 0;}ll three(ll x) {    ll y = pow(x, 1.0 / 3) + eps;    if (y * y * y == x) {        bool flag = 0;        for (int i = 1; i <= cnt; i++) {            if (y == p[i].x) {                flag = 1;                p[i].num += 3;                break;            }        }        if (!flag) {            p[++cnt].x = y;            p[cnt].num = 3;        }        return 1;    }    return 0;}ll two(ll x) {    ll y = pow(x, 1.0 / 2) + eps;    if (y * y == x) {        bool flag = 0;        for (int i = 1; i <= cnt; i++) {            if (y == p[i].x) {                flag = 1;                p[i].num += 2;                break;            }        }        if (!flag) {            p[++cnt].x = y;            p[cnt].num = 2;        }        return 1;    }    return 0;}void add_a(ll x, ll num) {    bool flag = 0;    for (int i = 1; i <= cur; i++) {        if (x == a[i].x) {            flag = 1;            a[i].num += num;            break;        }    }    if (!flag) {        a[++cur].x = x;        a[cur].num = num;    }}void add_p(ll x, ll num) {    bool flag = 0;    for (int i = 1; i <= cnt; i++) {        if (x == p[i].x) {            flag = 1;            p[i].num += num;            break;        }    }    if (!flag) {        p[++cnt].x = x;        p[cnt].num += num;    }}void try1(int i) {    bool flag = 0;    for (int j = i + 1; j <= cur; j++) {        ll y = __gcd(a[i].x, a[j].x);        if (y > 1) {            flag = 1;            add_p(y, a[i].num);            add_p(a[i].x / y, a[i].num);            break;        }    }    if (!flag) {        ans *= (a[i].num + 1);        ans %= mod;        ans *= (a[i].num + 1);        ans %= mod;    }}void solve() {    for (int i = 1; i <= cur; i++) {        bool flag = 0;        for (int j = 1; j <= cnt; j++) {            if (a[i].x % p[j].x == 0) {                add_p(p[j].x, a[i].num);                add_p(a[i].x / p[j].x, a[i].num);                flag = 1;                break;            }        }        if (!flag) {            try1(i);        }    }}void cal() {    for (int i = 1; i <= cnt; i++) {        ans *= (p[i].num + 1);        ans %= mod;    }    printf("%lld\n", ans);}int main() {    scanf("%d", &n);    for (int i = 1; i <= n; i++) {        ll x;        scanf("%lld", &x);        if (four(x) || three(x) || two(x)) {            continue;        } else {            add_a(x, 1);        }    }    solve();    cal();    return 0;}
   

代码解释：

通过上述步骤，代码能够高效地解决问题。

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