高斯消元样例全过，但0pts，悲

#include<iostream>
#include<cmath>
using namespace std;
typedef long long LL;
const int N = 2e3 + 10;
const LL mod = 998244353;
const double eps = 1e-8;
LL n, k, x[N], y[N], a[N][N];
LL qmi(LL a, LL k) {
	LL res = 1;
	while (k) {
		if (k & 1) res = res * a % mod;
		a = a * a % mod;
		k >>= 1;
	}
	return res;
}
LL f(LL x) {
	LL res = 0;
	for (int i = 0; i < n; ++i) {
		res = (res + a[i][n] * qmi(x, i) % mod) % mod;
	}
	return res;
}
void gauss(void) {
    int c, r;
    for (c = 0, r = 0; c < n; c++) {
        int t = r;
        for (int i = r; i < n; i++)   
            if (fabs(a[i][c]) > fabs(a[t][c]))
                t = i;
        if (fabs(a[t][c]) < eps) continue;

        for (int i = c; i <= n; i++) swap(a[t][i], a[r][i]);    
		for (int i = n; i >= c; i--) a[r][i] /= a[r][c];   
		for (int i = r + 1; i < n; i++)
			if (fabs(a[i][c]) > eps)
				for (int j = n; j >= c; j--)
					a[i][j] -= (a[r][j] * a[i][c]) % mod;

        r++;
    }
    for (int i = n - 1; i >= 0; i--)
        for (int j = i + 1; j < n; j++)
			a[i][n] -= (a[i][j] * a[j][n]) % mod;
}
int main(void) {
	cin >> n >> k;
	for (int i = 0; i < n; ++i) {
		cin >> x[i] >> y[i];
		for (int j = 0; j < n; ++j) {
			a[i][j] = qmi(x[i], j);
		}
		a[i][n] = y[i];
	}
	gauss();
	cout << f(k);
	return 0;
}