题解：离散化之后差分数组，注意左闭右闭区间的处理，将点和线段都抽象成点

#include <iostream>
#include <vector>
#include <set>
#include <algorithm>

using namespace std;

const int MAXN = 3000000 + 5;
const int INF = -2000000000;
vector<int> g;
int l[MAXN], r[MAXN], d[MAXN];

int get(int val) {
    return lower_bound(g.begin(), g.end(), val) - g.begin();
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n, x;
    cin >> n >> x;

    for (int i = 1; i <= n; ++i) {
        cin >> l[i] >> r[i];
        g.push_back(l[i] * 2);
        g.push_back(r[i] * 2 + 1);
    }

    sort(g.begin(), g.end());
    g.erase(unique(g.begin(), g.end()), g.end());

    for (int i = 1; i <= n; ++i) {
        d[get(l[i] * 2)]++;
        d[get(r[i] * 2 + 1)]--;
    }

    for (int i = 1; i < MAXN; ++i) {
        d[i] += d[i - 1];
    }

    vector<int> res;
    int last = INF;

    for (int i = 0; i < g.size(); ++i) {
        if (d[i] >= x) {
            if (last == INF) {
                last = (g[i] & 1) != 0 ? (g[i] - 1) / 2 : g[i] / 2;
            }
        } else {
            if (last == INF) continue;
            if (g[i] & 1) {
                res.push_back(last);
                res.push_back((g[i] - 1) / 2);
            } else {
                res.push_back(last);
                res.push_back(g[i] / 2);
            }
            last = INF;
        }
    }

    cout << res.size() / 2 << '\n';
    for (int i = 0; i < res.size(); i += 2) {
        cout << res[i] << " " << res[i + 1] << '\n';
    }

    return 0;
}