设取模前的答案为 $f(n)$。

打表可知，当 $n \ge 120$ 时，$f(n)\equiv 40176944 \pmod {10^8+7}$。

根据以上性质，我们直接对 $n>1000$ 的情况按照 $n=1000$ 暴力计算即可。

（实际上，这一性质是可以严格证明的，过程如下：）

$$\begin{aligned} x \otimes 119 &=\dfrac{119^2+2023x}{833+119x} \\ &=\dfrac{14161+2023x}{833+119x} \\ &=\dfrac{2023(7+x)}{119(7+x)} \\ &=\dfrac{2023}{119} \\ &=17 \end{aligned} $$$$\begin{aligned} f(n) &=n \otimes (n-1) \otimes (n-2) \otimes \ldots \otimes 1 \\ &=((n \otimes (n-1) \otimes (n-2) \otimes \ldots \otimes 120) \otimes 119) \otimes 118 \otimes 117 \otimes \ldots \otimes 1 \\ &=17 \otimes 118 \otimes 117 \otimes \ldots \otimes 1 \\ &\equiv 40176944 \pmod {10^8+7} \end{aligned} $$

#include <bits/stdc++.h>
using namespace std;
const int P = 1e8 + 7;
int read()
{
    int x = 0;
    char c = getchar_unlocked();
    while (c < '0') c = getchar_unlocked();
    while (c >= '0')
    {
        x = x * 10 + c - '0';
        c = getchar_unlocked();
        x = min(x, 1000);
    }
    return x;
}
int divs(int a, int b, int p = P)
{
	if (b % a == 0) return b / a;
	int x = divs(p % a, a - b % a, a);
	return (1ll * x * p + b) / a;
}
long long calc(long long x, long long y)
{
    return divs((833 + x * y) % P, (y * y + 2023 * x) % P);
}
signed main()
{
    freopen("calc.in", "r", stdin);
    freopen("calc.out", "w", stdout);
    int T = read();
    while (T--)
    {
        int n = read();
        int res = n;
        for (int i = n - 1; i >= 1; i--)
            res = calc(res, i);
        cout << res << endl;
    }
    return 0;
}