中国剩余定理模板

using ll = __int128;

template <typename T>
inline void rd(T &data) {
    T x = 0, flag = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch=='-') flag = flag * -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    data = x * flag;
}
template <typename T>
void wt(T x) {
     if (x < 0) putchar('-'), x = -x;
     if (x > 9) wt(x / 10);
     putchar(x % 10 + '0');
}

int n;
struct CRT {
    constexpr static int N = 30;
    ll a[N] {}, b[N] {};

    CRT() = default;
    CRT(int n) {
        for (int i = 1; i <= n; ++i) {
            rd(a[i]), rd(b[i]);
        }
    }
    ll exgcd(ll a, ll b, ll &x, ll &y) {
        if (!b) {
            x = 1, y = 0;
            return a;
        }
        ll d = exgcd(b, a % b, y, x);
        y -= a / b * x;
        return d;
    }
    ll getCRT() {  // 中国剩余定理
        ll M = 1;
        for (int i = 1; i <= n; i++) {
            M *= a[i];
        }
        ll res = 0;
        for (int i = 1; i <= n; i++) {
            ll Mi = M / a[i];
            ll x, y;
            exgcd(Mi, a[i], x, y);
            x = (x + a[i]) % a[i];
            res += b[i] * Mi * x;
        }
        return res % M;
    }
    ll getEXCRT() {  // 扩展中国剩余定理
        bool flag = true;
        ll a1 = a[1], b1 = b[1];
        for (int i = 2; i <= n; i++) {
            ll a2 = a[i], b2 = b[i], k1, k2;
            ll d = exgcd(a1, a2, k1, k2);
            if ((b2 - b1) % d) {
                flag = false;
                break;
            }
            k1 *= (b2 - b1) / d;
            k1 = (k1 % (a2 / d) + a2 / d) % (a2 / d);
            b1 = a1 * k1 + b1;
            a1 = a1 / d * a2;
            if (a1 < 0) {
                a1 = -a1;
            }
        }
        if (!flag) {
            return -1;
        } else {
            return (b1 % a1 + a1) % a1;
        }
    }
};