#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#define int long long
using namespace std;
const int maxn = 4e5 + 5;
int T[maxn], P[maxn], lenx, leny, n, res, KKIO, f[maxn];
struct ASK {
	int l, r, s, t;
}Q[maxn];
struct node {
	int L, R, sum;
	node operator + (const node &rhs) const {
		node res;
		res.L = L + rhs.L;
		res.R = R + rhs.R;
		res.sum = sum + rhs.sum + R * rhs.L;
		return res;
	}
}tr[maxn << 2];
vector<pair<int, int>> vec[maxn];
int ls(int p) {
	return p << 1;
}
int rs(int p) {
	return p << 1 | 1;
}
void build(int p, int l, int r) {
	tr[p].L = tr[p].R = tr[p].sum = 0;
	if(l == r) return;
	int mid = (l + r) >> 1;
	build(ls(p), l, mid);
	build(rs(p), mid + 1, r);
}
void update(int p, int l, int r, int nl, int op, int K) {
	if(l == r) {
		if(op) tr[p].R += K;
		else tr[p].L += K;
		return;
	}
	int mid = (l + r) >> 1;
	if(nl <= mid) update(ls(p), l, mid, nl, op, K);
	else update(rs(p), mid + 1, r, nl, op, K);
	tr[p] = tr[ls(p)] + tr[rs(p)];
}
int query(int p, int l, int r, int nl, int nr, int op) {
	if(nl > nr) return 0;
	if(nl <= l && r <= nr) return op ? tr[p].R : tr[p].L;
	int mid = (l + r) >> 1, res = 0;
	if(nl <= mid) res += query(ls(p), l, mid, nl, nr, op);
	if(nr > mid) res += query(rs(p), mid + 1, r, nl, nr, op);
	return res;
}
node query2(int p, int l, int r, int nl, int nr) {
	if(nl > nr) return {0, 0, 0};
	if(nl <= l && r <= nr) return tr[p];
	int mid = (l + r) >> 1;
	if(nl > mid) return query2(rs(p), mid + 1, r, nl, nr);
	if(nr <= mid) return query2(ls(p), l, mid, nl, nr);
	return query2(ls(p), l, mid, nl, nr) + query2(rs(p), mid + 1, r, nl, nr);
}
/*
6
4 9 5 6
5 8 4 7
6 7 8 10
3 10 3 11
2 12 1 12
1 11 2 9
5
6 10 1 3
2 7 7 10
8 9 2 5
3 4 4 9
1 5 6 8
*/
void solve() {
	int nowcnt = 0;
	for(int i = 1;i <= n;i++) vec[Q[i].l].push_back(make_pair(1, i)), vec[Q[i].r + 1].push_back(make_pair(-1, i));
	for(int i = 1;i <= lenx + 1;i++) {
		sort(vec[i].begin(), vec[i].end());
		for(pair<int, int> t : vec[i]) {
			int id = t.second;
			if(t.first == -1) {
				update(1, 1, leny, Q[id].s, 0, -1);
				update(1, 1, leny, Q[id].t, 1, -1);
				nowcnt --;
				continue;
			}
			int X = nowcnt - query(1, 1, leny, 1, Q[id].s - 1, 1) - query(1, 1, leny, Q[id].t + 1, leny, 0);
//			cout << t.second << " " << nowcnt << " " << query(1, 1, leny, 1, Q[id].s - 1, 1) << " " << query(1, 1, leny, Q[id].t + 1, leny, 0) << '\n';
			int Y = query2(1, 1, leny, Q[id].s, Q[id].t).sum;
			KKIO += X * (X - 1) / 2 - Y;
//			cout << X * (X - 1) / 2 - Y << '\n';
			nowcnt ++;
			update(1, 1, leny, Q[id].s, 0, 1);
			update(1, 1, leny, Q[id].t, 1, 1);
		}
		vec[i].clear();
	}
}
int a[maxn], b[maxn], c[maxn], d[maxn], belong[maxn], L[maxn], R[maxn];
bitset<maxn> A[505], B[505], C[505], D[505], E, EA, EB, EC, ED;
void bit() {
	for(int i = 1;i <= n;i++) a[i] = b[i] = c[i] = d[i] = i;
	sort(a + 1, a + 1 + n, [&](int x, int y) {return Q[x].r > Q[y].r;});
	sort(b + 1, b + 1 + n, [&](int x, int y) {return Q[x].l < Q[y].l;});
	sort(c + 1, c + 1 + n, [&](int x, int y) {return Q[x].t > Q[y].t;});
	sort(d + 1, d + 1 + n, [&](int x, int y) {return Q[x].s < Q[y].s;});
	int blc = sqrt(n);
	for(int i = 1;i <= n;i++) belong[i] = i / blc + (i % blc != 0);
	for(int i = 1;i <= belong[n];i++) L[i] = R[i - 1] + 1, R[i] = min(i * blc, n);
	for(int i = 1;i <= belong[n];i++) {
		A[i] = A[i - 1], B[i] = B[i - 1], C[i] = C[i - 1], D[i] = D[i - 1];
		for(int j = L[i];j <= R[i];j++) A[i][a[j]] = B[i][b[j]] = C[i][c[j]] = D[i][d[j]] = 1;
	}
	for(int i = 1;i <= n;i++) {
		int l = 1, r = n, aa = 0, bb = 0, cc = 0, dd = 0;
		while(l <= r) {
			int mid = (l + r) >> 1;
			if(Q[a[mid]].r >= Q[i].l) l = mid + 1, aa = mid;
			else r = mid - 1;
		}
		l = 1, r = n;
		while(l <= r) {
			int mid = (l + r) >> 1;
			if(Q[b[mid]].l <= Q[i].r) l = mid + 1, bb = mid;
			else r = mid - 1;
		}
		l = 1, r = n;
		while(l <= r) {
			int mid = (l + r) >> 1;
			if(Q[c[mid]].t >= Q[i].s) l = mid + 1, cc = mid;
			else r = mid - 1;
		}
		l = 1, r = n;
		while(l <= r) {
			int mid = (l + r) >> 1;
			if(Q[d[mid]].s <= Q[i].t) l = mid + 1, dd = mid;
			else r = mid - 1;
		}
		EA = A[belong[aa] - 1];
		for(int j = L[belong[aa]];j <= aa;j++) EA[a[j]] = 1;
		EB = B[belong[bb] - 1];
		for(int j = L[belong[bb]];j <= bb;j++) EB[b[j]] = 1;
		EC = C[belong[cc] - 1];
		for(int j = L[belong[cc]];j <= cc;j++) EC[c[j]] = 1;
		ED = D[belong[dd] - 1];
		for(int j = L[belong[dd]];j <= dd;j++) ED[d[j]] = 1;
		E = (EA & EB & EC & ED);
		f[i] = E.count() - 1;
	}
}
signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	cin >> n;
	for(int i = 1;i <= n;i++) {
		cin >> Q[i].l >> Q[i].r >> Q[i].s >> Q[i].t;
		T[++ lenx] = Q[i].l, T[++ lenx] = Q[i].r;
		P[++ leny] = Q[i].s, P[++ leny] = Q[i].t;
	}
	sort(T + 1, T + 1 + lenx), sort(P + 1, P + 1 + leny);
	lenx = unique(T + 1, T + 1 + lenx) - T - 1, leny = unique(P + 1, P + 1 + leny) - P - 1;
	for(int i = 1;i <= n;i++) {
		Q[i].l = lower_bound(T + 1, T + 1 + lenx, Q[i].l) - T;
		Q[i].r = lower_bound(T + 1, T + 1 + lenx, Q[i].r) - T;
		Q[i].s = lower_bound(P + 1, P + 1 + leny, Q[i].s) - P;
		Q[i].t = lower_bound(P + 1, P + 1 + leny, Q[i].t) - P;
	}
	solve();
	bit();
	int ans1 = 0, ans2 = 0;
	for(int i = 1;i <= n;i++) {
		ans1 += f[i] * (n - 2), ans2 += f[i] * (f[i] - 1) / 2;
	}
	res = (ans1 - ans2 * 2) / 2;
	res += KKIO;
	cout << n * (n - 1) * (n - 2) / 6 - res << '\n'; 
}