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;
vector<uint32_t> gen_permutation(uint32_t n, uint64_t k) {
  vector<uint32_t> factoradic;
  factoradic.reserve(n);
  for (uint32_t p = 1; 0 < k && p <= n; k /= p++) {
    factoradic.emplace_back(static_cast<uint32_t>(k % p));
  }
  deque<uint32_t> sequence(factoradic.size());
  iota(begin(sequence), end(sequence),
       n - static_cast<uint32_t>(factoradic.size()) + 1);
  vector<uint32_t> permutation;
  permutation.reserve(n);
  permutation.resize(n - static_cast<uint32_t>(factoradic.size()));
  iota(begin(permutation), end(permutation), 1);
  while (!factoradic.empty()) {
    auto it = next(begin(sequence), factoradic.back());
    permutation.emplace_back(*it);
    sequence.erase(it);
    factoradic.pop_back();
  }
  return permutation;
}
void update_permutation(vector<uint32_t>& permutation, uint64_t k) {
  vector<uint32_t> factoradic;
  factoradic.reserve(permutation.size());
  for (uint32_t p = 1; 0 < k && p <= permutation.size(); k /= p++) {
    factoradic.emplace_back(static_cast<uint32_t>(k % p));
  }
  deque<uint32_t> sequence(factoradic.size());
  iota(begin(sequence), end(sequence),
       static_cast<uint32_t>(permutation.size() - factoradic.size()) + 1);
  permutation.erase(
      next(begin(permutation), permutation.size() - factoradic.size()),
      end(permutation));
  while (!factoradic.empty()) {
    auto it = next(begin(sequence), factoradic.back());
    permutation.emplace_back(*it);
    sequence.erase(it);
    factoradic.pop_back();
  }
}
uint64_t arithmetic_progression(vector<uint32_t> const& numbers, int lh,
                                int rh) {
  return static_cast<uint64_t>(numbers[lh] + numbers[rh]) * (rh - lh + 1) / 2;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int N, Q, T, L, R, X;
  cin >> N >> Q;
  uint64_t k = 0;
  vector<uint32_t> numbers(N);
  iota(begin(numbers), end(numbers), 1);
  const int R_MAX = N - 14;
  for (int q = 0; q < Q; ++q) {
    cin >> T;
    switch (T) {
      case 1:
        cin >> L >> R;
        if (R < R_MAX) {
          cout << arithmetic_progression(numbers, L - 1, R - 1) << '\n';
        } else if (L < R_MAX) {
          cout << accumulate(next(begin(numbers), R_MAX),
                             next(begin(numbers), R),
                             arithmetic_progression(numbers, L - 1, R_MAX - 1))
               << '\n';
        } else {
          cout << accumulate(next(begin(numbers), L - 1),
                             next(begin(numbers), R), 0ull)
               << '\n';
        }
        break;
      case 2:
        cin >> X;
        update_permutation(numbers, k += X);
        break;
    }
  }
  return 0;
}