Problem: A permutation is a sequence of integers from 1 to n of length n containing each number exactly once. For example, [1], [4, 3, 5, 1, 2], [3, 2, 1] — are permutations, and [1, 1], [4, 3, 1], [2, 3, 4] — no.

Permutation a is lexicographically smaller than permutation b (they have the same length n), if in the first index i in which they differ, a[i] < b[i]. For example, the permutation [1, 3, 2, 4] is lexicographically smaller than the permutation [1, 3, 4, 2], because the first two elements are equal, and the third element in the first permutation is smaller than in the second.

The next permutation for a permutation a of length n — is the lexicographically smallest permutation b of length n that lexicographically larger than a. For example: 

  * for permutation [2, 1, 4, 3] the next permutation is [2, 3, 1, 4]; 
  * for permutation [1, 2, 3] the next permutation is [1, 3, 2]; 
  * for permutation [2, 1] next permutation does not exist. 



You are given the number n — the length of the initial permutation. The initial permutation has the form a = [1, 2, …, n]. In other words, a[i] = i (1 ≤ i ≤ n).

You need to process q queries of two types: 

  * 1 l r: query for the sum of all elements on the segment [l, r]. More formally, you need to find a[l] + a[l + 1] + … + a[r]. 
  * 2 x: x times replace the current permutation with the next permutation. For example, if x=2 and the current permutation has the form [1, 3, 4, 2], then we should perform such a chain of replacements [1, 3, 4, 2] → [1, 4, 2, 3] → [1, 4, 3, 2]. 



For each query of the 1-st type output the required sum.

Input

The first line contains two integers n (2 ≤ n ≤ 2 ⋅ 10^5) and q (1 ≤ q ≤ 2 ⋅ 10^5), where n — the length of the initial permutation, and q — the number of queries.

The next q lines contain a single query of the 1-st or 2-nd type. The 1-st type query consists of three integers 1, l and r (1 ≤ l ≤ r ≤ n), the 2-nd type query consists of two integers 2 and x (1 ≤ x ≤ 10^5).

It is guaranteed that all requests of the 2-nd type are possible to process.

Output

For each query of the 1-st type, output on a separate line one integer — the required sum.

Example

Input


4 4
1 2 4
2 3
1 1 2
1 3 4


Output


9
4
6

Note

Initially, the permutation has the form [1, 2, 3, 4]. Queries processing is as follows: 

  1. 2 + 3 + 4 = 9; 
  2. [1, 2, 3, 4] → [1, 2, 4, 3] → [1, 3, 2, 4] → [1, 3, 4, 2]; 
  3. 1 + 3 = 4; 
  4. 4 + 2 = 6 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
using lpair = std::pair<ll, ll>;
const int _ = []() {
  std::cin.tie(nullptr);
  std::ios::sync_with_stdio(false);
  std::cout << std::fixed << std::setprecision(10);
  return 0;
}();
template <typename value_t>
void resize(value_t& v, const value_t& val) {
  v = val;
}
template <typename vec_t, typename value_t, typename... arg_t>
void resize(std::vector<vec_t>& v, const value_t& val, int size, arg_t... arg) {
  v.resize(size);
  for (auto& c : v) resize(c, val, arg...);
}
template <typename A, typename B>
long long imin(A a, B b) {
  return min(static_cast<long long>(a), static_cast<long long>(b));
}
template <typename A, typename B>
long long imax(A a, B b) {
  return max(static_cast<long long>(a), static_cast<long long>(b));
}
template <typename A, typename B>
void chmin(A& a, const B& b) {
  a = min(a, static_cast<A>(b));
};
template <typename A, typename B>
void chmax(A& a, const B& b) {
  a = max(a, static_cast<A>(b));
};
template <typename X>
class segtree {
 public:
  using func_xx = std::function<X(X, X)>;
  long long N;
  func_xx fxx;
  X ex;
  std::vector<X> dat;
  bool changed;
  segtree() {}
  segtree(long long n, func_xx fxx, const X& ex)
      : fxx(fxx), ex(ex), changed(false) {
    if (n == 1) ++n;
    int x = 1;
    while (x < n) x *= 2;
    N = x;
    dat.resize(2 * N - 1, ex);
  }
  const segtree<X>& operator=(const segtree& source) {
    N = source.N;
    fxx = source.fxx;
    ex = source.ex;
    dat = source.dat;
    changed = source.changed;
    return *this;
  }
  void update(int k) {
    dat[k] = fxx(dat[2 * k + 1], dat[2 * k + 2]);
    if (k != 0) update((k - 1) / 2);
  }
  void build() {
    for (int i = N - 2; i >= 0; --i)
      dat[i] = fxx(dat[2 * i + 1], dat[2 * i + 2]);
    changed = false;
  }
  void set(int i, const X& x, bool update = true) {
    dat[N - 1 + i] = x;
    if (update)
      this->update((N - 2 + i) / 2);
    else
      changed = true;
  }
  const X& elem(int i) { return dat[N - 1 + i]; }
  X query_sub(int a, int b, int k, int l, int r) {
    if (r <= a || b <= l) {
      return ex;
    } else if (a <= l && r <= b) {
      return dat[k];
    } else {
      return fxx(query_sub(a, b, 2 * k + 1, l, (l + r) / 2),
                 query_sub(a, b, 2 * k + 2, (l + r) / 2, r));
    }
  }
  X query(int a, int b) {
    if (changed) build();
    return query_sub(a, b, 0, 0, N);
  }
};
int main() {
  ll n, q;
  cin >> n >> q;
  segtree<ll> seg{n, [](ll a, ll b) { return a + b; }, 0};
  for (ll i = 0; i < n; ++i) {
    seg.set(i, i + 1, false);
  }
  ll cum = 0;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int l, r;
      cin >> l >> r;
      --l, --r;
      cout << seg.query(l, r + 1) << endl;
    } else {
      ll x;
      cin >> x;
      cum += x;
      ll size = imin(n, 14);
      ll frac = 1;
      for (ll i = 2; i < size; ++i) {
        frac *= i;
      }
      vector<bool> used(size, false);
      ll tmp = cum;
      for (ll i = size - 1; i > -1; --i) {
        ll j = 0, cnt = 0, ind = tmp / frac;
        for (;; ++j) {
          if (used[j]) continue;
          if (cnt == ind) {
            used[j] = true;
            break;
          }
          ++cnt;
        }
        seg.set(n - i - 1, n - size + j + 1);
        if (i == 0) break;
        tmp %= frac;
        frac /= i;
      }
    }
  }
  return 0;
}