#include <iostream>
#include <cmath>
#include <vector>
#include <tuple>
#include <algorithm>

using namespace std;

using ll = long long;
using vi = vector<int>;
using v2i = vector<vi>;
using t4i = tuple<int, int, int, int>;

constexpr int X = 1e9;

t4i exgcd(int a, int b) {
    if (b == 0) {
        return {1, 0, 0, 1};
    }

    auto [k1, k2, k3, k4] = exgcd(b, a % b);
    return {k2, k1 - a / b * k2, k4, k3 - a / b * k4};
}

int main() {
    int k, q;
    cin >> k >> q;
    int n = log(X) / log(k) + 1;

    v2i base(n);

    while (q--) {
        int op;
        ll x;
        cin >> op >> x;

        if (op == 1) {
            int xd[n];
            for (int i = 0; i < n; i++) {
                xd[i] = x % k;
                x /= k;
            }

            for (int i = n - 1; i >= 0; i--) {
                if (xd[i] == 0) {
                    continue;
                }

                auto [k1, k2, k3, k4] = exgcd(xd[i], base[i].empty() ? k : base[i][i]);

                if (base[i].empty()) {
                    base[i] = vi(n);
                }

                int t[n];
                copy(base[i].begin(), base[i].end(), t);

                for (int j = 0; j < n; j++) {
                    base[i][j] = ((1ll * k1 * xd[j] + 1ll * k2 * t[j]) % k + k) % k;
                }
                for (int j = 0; j < n; j++) {
                    xd[j] = ((1ll * k3 * xd[j] + 1ll * k4 * t[j]) % k + k) % k;
                }
            }
        } else if (op == 2) {
            x--;

            int xd[n];
            for (int i = 0; i < n; i++) {
                if (!base[i].empty()) {
                    xd[i] = x % (k / base[i][i]);
                    x /= k / base[i][i];
                }
            }

            if (x > 0) {
                cout << -1 << endl;
                continue;
            }

            int ans[n];
            fill(ans, ans + n, 0);
            for (int i = n - 1; i >= 0; i--) {
                if (!base[i].empty()) {
                    int t = ((xd[i] - ans[i] / base[i][i]) % k + k) % k;
                    for (int j = 0; j < n; j++) {
                        ans[j] = (ans[j] + 1ll * t * base[i][j]) % k;
                    }
                }
            }

            ll t = 0;
            for (int i = n - 1; i >= 0; i--) {
                t = t * k + ans[i];
            }

            cout << t << endl;
        } else {
            int xd[n];
            for (int i = 0; i < n; i++) {
                xd[i] = x % k;
                x /= k;
            }

            if (x > 0) {
                xd[n-1] = k;
            }

            int r[n];
            fill(r, r + n, 0);

            ll dp[n+1][2];
            fill(dp[0], dp[n+1], 0);
            dp[n][1] = 1;
            for (int i = n - 1; i >= 0; i--) {
                int t = base[i].empty() ? k : base[i][i];

                dp[i][0] = dp[i+1][0] * (k / t);
                dp[i][0] += dp[i+1][1] * (xd[i] / t + (r[i] % t < xd[i] % t));

                if (xd[i] < k && r[i] % t == xd[i] % t) {
                    dp[i][1] = dp[i+1][1];

                    if (!base[i].empty()) {
                        int k1 = ((xd[i] - r[i]) / t + k) % k;
                        for (int j = 0; j < n; j++) {
                            r[j] = (r[j] + 1ll * k1 * base[i][j]) % k;
                        }
                    }
                }
            }

            cout << dp[0][0] + dp[0][1] << endl;
        }
    }

    return 0;
}