#include <bits/stdc++.h>
#define cerr cout << "in " << __LINE__ << "\t: "
using namespace std;
const int mod = 998244353;
struct mint {
	int x;
	mint() : x(0) {}
	mint(long long y, bool flag = 0) {
		if (flag) x = y;
		else x = (y % mod + mod) % mod;
	}
	friend const mint ksm(mint a, long long b);
	const mint inv() {return ksm(*this, mod - 2);}
};
bool operator == (const mint a, const mint b) {return a.x == b.x;}
bool operator != (const mint a, const mint b) {return a.x != b.x;}
bool operator < (const mint a, const mint b) {return a.x < b.x;}
int operator ! (const mint a) {return !a.x;}
const mint operator + (const mint a, const mint b) {
	mint res(a.x + b.x, 1);
	if (res.x >= mod) res.x -= mod;
	return res;
}
mint& operator += (mint &a, const mint b) {
	a.x += b.x;
	if (a.x >= mod) a.x -= mod;
	return a;
}
const mint operator - (const mint a, const mint b) {
	mint res(a.x - b.x, 1);
	if (res.x < 0) res.x += mod;
	return res;
}
mint& operator -= (mint &a, const mint b) {
	a.x -= b.x;
	if (a.x < 0) a.x += mod;
	return a;
}
const mint operator * (const mint a, const mint b) {
	return mint((long long)a.x * b.x % mod, 1);
}
mint& operator *= (mint &a, const mint b) {
	a.x = (long long)a.x * b.x % mod;
	return a;
}
const mint ksm(mint a, long long b) {
	mint res(1, 1);
	for (; b; a *= a, b >>= 1)
		if (b & 1) res *= a;
	return res;
}
const mint operator / (const mint a, const mint b) {
	return a * ksm(b, mod - 2);
}
mint& operator /= (mint &a, const mint b) {
	a = a * ksm(b, mod - 2);
	return a;
}
ostream& operator << (ostream &out, const mint a) {
	return out << a.x;
}
istream& operator >> (istream &in, mint &a) {
	long long y;
	in >> y, a = mint(y);
	return in;
}
int n, t[200010], a[200010], b[200010], p[200010];
signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	cin >> n;
	for (int i = 1; i <= n; i++)
		cin >> t[i] >> a[i] >> b[i], p[i] = i;
	sort(p + 1, p + n + 1, [&](int x, int y) {
		return t[y] * b[y] * (b[x] - a[x]) > t[x] * b[x] * (b[y] - a[y]);
	});
	mint sum = 1, ans = 0;
	for (int i = n; i >= 1; i--)
		sum *= mint(b[p[i]]) / a[p[i]], ans += t[p[i]] * sum;
	cout << ans << "\n";
	return 0;
}