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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  vector<int> A(n + 1);
  for (int i = 1; i <= n; ++i) {
    A[i] = i;
  }
  vector<long long> S(n + 1);
  for (int i = 1; i <= n; ++i) {
    S[i] = S[i - 1] + A[i];
  }
  while (q--) {
    int cmd;
    cin >> cmd;
    if (cmd == 1) {
      int l, r;
      cin >> l >> r;
      cout << S[r] - S[l - 1] << endl;
    } else {
      int x;
      cin >> x;
      int first_changed = n;
      while (x > 0) {
        int i = n;
        int fact = 1;
        while (i - 1 >= 0 && A[i - 1] < A[i] &&
               fact * (n - (i - 1) + 1) - 1 <= x) {
          fact *= (n - (i - 1) + 1);
          --i;
        }
        if (fact == 1) {
          i = n;
          while (A[i - 1] > A[i]) {
            --i;
          }
          assert(A[i - 1] < A[i]);
          int j = n;
          while (A[j] < A[i - 1]) {
            --j;
          }
          assert(A[j] > A[i - 1]);
          assert(j >= i);
          swap(A[i - 1], A[j]);
          reverse(A.begin() + i, A.end());
          first_changed = min(first_changed, i - 1);
          --x;
        } else {
          assert(i < n);
          reverse(A.begin() + i, A.end());
          x -= (fact - 1);
          first_changed = min(first_changed, i);
        }
      }
      for (int i = first_changed; i <= n; ++i) {
        S[i] = S[i - 1] + A[i];
      }
    }
  }
  return 0;
}