#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=a;i<b;i++)




typedef long double ld;
typedef long long ll;
int N;
vector<ld> vec[3];
ld EPS = 1e-18;
//-----------------------------------------------------------------------------------
ll count() {
    ll ans = 0;

    sort(vec[2].begin(), vec[2].end());

    for (ld a : vec[0]) for (ld b : vec[1]) {

        int cnt = 0, _cnt = 0;

        // Naive
        /*for (ld c : vec[2]) {
            ld sum = a + b + c;
            if (abs(1 - sum) < EPS) continue;

            ld ab = a + b;
            ld bc = b + c;
            ld ac = a + c;

            if (ab <= 1 && bc <= 1 && ac <= 1) _cnt++;
        }*/

        // Optimized
        ld ab = a + b;
        if (1 < ab) continue;

        int minus = 0;
        // c = 1 - a - b があるか探す
        int lo = lower_bound(vec[2].begin(), vec[2].end(), 1 - a - b - EPS) - vec[2].begin();
        int up = upper_bound(vec[2].begin(), vec[2].end(), 1 - a - b + EPS) - vec[2].begin();
        if (lo != up) minus = -1;
        // c <= min(1-b,1-a) の個数を探す
        up = upper_bound(vec[2].begin(), vec[2].end(), min(1 - b, 1 - a) + EPS) - vec[2].begin();
        cnt = up + minus;

        //if (_cnt != cnt) printf("eer!\n");

        ans += cnt;
    }

    return ans;
}
//-----------------------------------------------------------------------------------
int main() {
    cin >> N;
    rep(i, 0, 3) vec[i].push_back(0);
    rep(i, 0, N) {
        int P, A, B;
        scanf("%d%d%d", &P, &A, &B);
        vec[P].push_back((ld)B / (ld)(A + B));
    }

    cout << count() << endl;
}