在洛谷上过了，为什么在这里会RE?

#include <iostream>
#include <cstdio>
typedef long long LL;
using namespace std;

void RD() {}
template<typename T, typename... U> void RD(T &x, U&... arg) {
    x = 0; int f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    x *= f; RD(arg...);
}

const LL N = 2e7 + 5, mod = 1e9 + 7;

#define PRINT(x) cout << #x << " = " << x << "\n"
#define LF(i, __l, __r) for (int i = __l; i <= __r; i++)
#define RF(i, __r, __l) for (int i = __r; i >= __l; i--)

LL n, cnt;
LL a, b, c, g[N], ans;
LL f[N], num[N], prime[N];
bool vis[N];

void get(int n) {
    f[1] = 1;
    LF(i, 2, n) {
        if (!vis[i]) prime[++cnt] = i, f[i] = 2, num[i] = 1;

        LF(j, 1, cnt) {
            int nx = i * prime[j];
            if (nx > n) break;
            vis[nx] = 1;

            if (i % prime[j] == 0) {
                num[nx] = num[i] + 1;
                f[nx] = f[i] / num[nx] * (num[nx] + 1);
                break; 
            }

            num[nx] = 1; f[nx] = f[i] * 2;
        }
    }
}

int main() {
    RD(n, a, b, c);
    RD(g[n]);
    RF(i, n - 1, 1) g[i] = (a * g[i + 1] % mod * g[i + 1] % mod + b * g[i + 1] % mod + c) % mod;
    get(n);
    LF(i, 1, n) f[i] += f[i - 1];
    LF(i, 1, n) ans = (ans + f[i - 1] * (n - i + 1) % mod * g[i] % mod + n * (n - i + 1) % mod * g[i] % mod) % mod;
    printf("%lld", ans);
    return 0;
}