ABC 414

void solve() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int N, L, R;
    cin >> N >> L >> R;

    ll ans = 0;
    rep(i, N) {
        int x, y;
        cin >> x >> y;
        if (x <= L && R <= y) ans++;
    }
    cout << ans << endl;
}

void solve() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int N;
    cin >> N;
    vector<pair<char, ll>> ps;
    ll tot = 0;
    rep(i, N) {
        char c;
        ll l;
        cin >> c >> l;
        ps.emplace_back(c, l);
        tot += l;
        if (tot > 100) {
            cout << "Too Long" << endl;
            return;
        }
    }

    for (auto [c, n] : ps) {
        rep(i, n) cout << c;
    }
    cout << endl;
}

void solve() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    ll A, N;
    cin >> A >> N;

    auto is_palindrome_a = [&](ll s) -> bool {
        vector<ll> memo;
        while (s) {
            memo.push_back(s % A);
            s /= A;
        }

        vll rev = memo;
        reverse(all(memo));
        return memo == rev;
    };

    ll ans = 0;
    // 1桁のとき
    rep2(i, 1, 10) {
        if (is_palindrome_a(i) && i <= N) ans += i;
    }

    // 2桁以上のときの数字
    for (ll i = 1; i <= (ll)1e6; i++) {
        string s, srev;
        s = to_string(i);
        srev = s;
        reverse(all(srev));

        // 偶数長
        {
            ll cand = stoll(s + srev);
            if (cand <= N && is_palindrome_a(cand)) {
                ans += cand;
            }
        }

        // 奇数長
        rep(i, 10) {
            string x;
            x.push_back('0' + i);
            ll cand = stoll(s + x + srev);
            if (cand <= N && is_palindrome_a(cand)) {
                ans += cand;
            }
        }
    }

    cout << ans << endl;
}

void solve() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    ll N, M;
    cin >> N >> M;
    vll X(N);
    rep(i, N) cin >> X[i];
    sort(all(X));

    vll dists;
    rep(i, N - 1) {
        dists.push_back(X[i + 1] - X[i]);
    }

    sort(all(dists));
    rep(i, M - 1) {
        dists.pop_back();
    }

    ll ans = accumulate(all(dists), 0ll);
    cout << ans << endl;
}