#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
using namespace std;

//segment tree

#define segL (1 << 20)
int segT[segL * 2];

void init() {
    for (int i = 0;i < segL * 2;i++)segT[i] = 0;
}

int c2(int X) {
    int res = 0;
    int t = X;
    if (t == 0)t = segL;
    while (1) {
        if (t % 2 == 0) {
            t /= 2;
            res++;
        }
        else {
            break;
        }
    }
    return res;
}

void update(int idx, long long X) {
    int i = idx + segL;
    segT[i] = X;
    i /= 2;
    while (i >= 1) {
        segT[i] = segT[2 * i] + segT[2 * i + 1];
        i /= 2;
    }
}

int getS(int L, int R) {
    int res = 0;
    int tL = L;
    while (tL != R) {
        int b = c2(tL);
        for (; b >= 0; b--) {
            if (tL + (1 << b) <= R) {
                int segIdx = (tL + segL) >> b;
                res += segT[segIdx];
                tL += (1 << b);
                break;
            }
        }
    }
    return res;
}

// segment tree

//無重複の数列に対して、転倒数を求める関数。
long long inversion(const vector<int> V) {
    init();
    long long res = 0;
    for (int i = 0;i < V.size();i++) {
        res += (long long)getS(V[i], segL);
        update(V[i], 1);
    }
    return res;
}

//端点が無重複の区間列に対して、包含関係をなす区間の組の数を求める関数。
long long count_included_relation(const vector<int> L, const vector<int> R) {
    init();
    vector<pair<int, int>>V;
    for (int i = 0;i < L.size();i++) {
        V.push_back(make_pair(L[i], R[i]));
    }
    sort(V.begin(), V.end());
    long long res = 0;
    for (int i = 0;i < V.size();i++) {
        res += (long long)getS(V[i].second, segL);
        update(V[i].second, 1);
    }
    return res;
}

int main() {
    int N;
    cin >> N;
    vector<int>L(N), R(N);
    set<int>check_;
    bool ng_ = 0;
    for (int i = 0;i < N;i++) {
        cin >> L[i] >> R[i];
        if (L[i] <= 0 || L[i] >= R[i] || 1'000'001 <= R[i])ng_ = 1;
        check_.insert(L[i]);check_.insert(R[i]);
    }
    if (check_.size() != 2 * N || ng_) {//制約違反検出
        int tmptmp1 = 1;
        int tmptmp0 = 0;
        N = tmptmp1 / tmptmp0;//Run Time Errorさせる
    }
    long long ans = 0;
    ans += count_included_relation(L, R);
    ans += inversion(R);
    reverse(L.begin(), L.end());
    ans += inversion(L);
    long long NC2 = ((long long)N * (long long)(N - 1))/(long long)2;
    ans -= NC2;
    ans /= (long long)2;
    cout << ans;
    return 0;
}