I have solved this problem by using an easy greedy algorithm.

MAIN: you need to use an idea of 2 pointers(of course you need to sort array) and take the rightest segment. Left part of segment need to be <= x(another pointer)

The solution below:

void solve() {
    int m;
    cin >> m;
    vector< pair<int, int> > pi;
    int a, b;
    cin >> a >> b;
    while (a || b) {
        pi.pb({a, b});
        cin >> a >> b;
    }
    sort(all(pi));
    int cnt = 0, j = 0;
    vector< pair<int, int> > ans;
    j = 0;
    int x = 0;
    for (; j < sz(pi) && x < m; j++) {
        int k = j;
        int mx = x;
        pair<int, int> par = {-INF, -INF};
        while (k < sz(pi) && pi[k].ff <= x) {
            if (mx < pi[k].ss) {
                mx = pi[k].ss;
                par = pi[k];
            }
            k++;
        }
        if (par.ff == -INF) {
            cout << "No solution\n";
            return;
        }
        x = mx;
        ans.pb(par);
        if (k > j)
            j = k - 1;
    }
    if (x < m) {
        cout << "No solution\n";
        return;
    }
    cout << sz(ans) << '\n';
    for (auto &i : ans) {
        cout << i.ff << ' ' << i.ss << '\n';
    }
}

always choose segments with rightest right points