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;
istream& inp = cin;
template <typename T>
vector<T> readV(int size) {
  vector<T> result;
  for (int i = 0; i < size; ++i) {
    T x;
    inp >> x;
    result.push_back(x);
  }
  return result;
}
template <typename T>
T divup(T a, T b) {
  if ((a % b) == 0) return a / b;
  return 1 + a / b;
}
const int64_t M = 5e+10;
vector<int64_t> fac{1};
int PSize = 0;
int64_t toFac(vector<int> p, int pos) {
  if (p.size() == pos) return 0;
  int64_t r = fac.at(p.size() - pos - 1) * p.at(pos) - 1;
  for (int i = pos + 1; i < p.size(); ++i)
    if (p.at(i) > p.at(pos)) --p.at(i);
  return r + toFac(p, pos + 1);
}
vector<int> fromFac(int64_t x) {
  vector<int> R(PSize);
  vector<bool> occ(PSize + 1, false);
  for (int i = 0; i < PSize; ++i) {
    int64_t u = x / fac.at(PSize - i - 1);
    int v = 1;
    while (occ.at(v)) ++v;
    for (int j = 0; j < u; ++j) {
      ++v;
      while (occ.at(v)) ++v;
    }
    R.at(i) = v;
    occ.at(v) = true;
    x = x % fac.at(PSize - i - 1);
  }
  return R;
}
void cv(vector<int> const& v) {
  for (int x : v) cout << x << " ";
  cout << endl;
}
int main() {
  assert(inp);
  int64_t f = 1;
  for (int64_t i = 1; f <= M * 100; ++i) {
    f *= i;
    fac.push_back(f);
  }
  PSize = 4;
  int n, q;
  inp >> n >> q;
  while (fac.at(PSize) <= M) ++PSize;
  PSize = min(PSize, n);
  const int64_t bound = n - PSize;
  int64_t perm = 0;
  for (int qn = 0; qn < q; ++qn) {
    int qt;
    inp >> qt;
    if (2 == qt) {
      int x;
      inp >> x;
      perm += x;
    } else {
      int64_t l, r;
      inp >> l >> r;
      --l;
      --r;
      auto end = fromFac(perm);
      int64_t sum = 0;
      if (l < bound) {
        const int64_t r0 = min(bound - 1, r) + 1;
        sum += r0 * (r0 + 1) / 2 - (l) * (l + 1) / 2;
        l = 0;
      } else
        l -= bound;
      r -= bound;
      if (r >= 0) {
        auto p = fromFac(perm);
        for (int i = l; i <= r; ++i) sum += p.at(i) + bound;
      }
      cout << sum << endl;
    }
  }
  return 0;
}