// Freezer — Greedy Approach O(N² log N)
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    int n;
    cin >> n;
    vector<int> s(n), m(n);
    for (int i = 0; i < n; i++) cin >> s[i] >> m[i];

    vector<bool> used(n, false);
    vector<pair<int,int>> result;
    int day = 1;

    while (true) {
        // Collect available meals: not used, quality >= 4, not expired
        vector<int> avail;
        for (int i = 0; i < n; i++)
            if (!used[i] && m[i] >= 4 && s[i] >= day)
                avail.push_back(i);

        // Sort by expiry (eat soonest-expiring first)
        sort(avail.begin(), avail.end(),
            [&](int a, int b){ return s[a] < s[b]; });

        // Try to find a valid pair
        bool found = false;
        for (int i = 0; i < (int)avail.size() && !found; i++)
            for (int j = i+1; j < (int)avail.size() && !found; j++)
                if (m[avail[i]] + m[avail[j]] >= 9) {
                    used[avail[i]] = used[avail[j]] = true;
                    result.push_back({avail[i]+1, avail[j]+1});
                    found = true;
                }
        if (!found) break;
        day++;
    }

    cout << result.size() << endl;
    for (auto& [a, b] : result)
        cout << a << " " << b << "\n";
    return 0;
}