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;
long long int arr1[50], arr2[50];
vector<long long int> vec1;
vector<long long int> vec2;
long long int stair[16];
int main() {
  cin.tie(0);
  stair[0] = 1;
  long long int n, q, c;
  for (int i = 1; i <= 15; i++) {
    stair[i] = stair[i - 1] * i;
  }
  while (cin >> n >> q) {
    if (n > 15) {
      c = 15;
    } else {
      c = n;
    }
    for (int i = 0; i < c; i++) {
      vec1.push_back(n - i);
    }
    vector<long long int>::iterator u;
    sort(vec1.begin(), vec1.end());
    long long int m = 0;
    for (int i = 0; i < q; i++) {
      int a;
      cin >> a;
      if (a == 1) {
        vec2 = vec1;
        long long int l, r;
        cin >> l >> r;
        if (r <= n - 15) {
          cout << ((r + 1) * r) / 2 - ((l - 1) * l) / 2 << "\n";
        } else if (l <= n - 15) {
          long long int ans = 0;
          ans = ans + ((n - 15 + 1) * (n - 15)) / 2 - ((l - 1) * l) / 2;
          for (int j = 15; j >= 1; j--) {
            long long int y = m / stair[j - 1];
            y = y % j + 1;
            arr1[16 - j] = vec2[y - 1];
            for (u = vec2.begin(); u != vec2.end(); u++) {
              if (*u == arr1[15 + 1 - j]) {
                vec2.erase(u);
                break;
              }
            }
          }
          l = 1;
          r = r - (n - 15);
          for (int j = l; j <= r; j++) {
            ans = ans + arr1[j];
          }
          cout << ans << "\n";
        } else {
          long long int ans = 0;
          vec2 = vec1;
          for (int j = c; j >= 1; j--) {
            long long int y = m / stair[j - 1];
            y = y % j + 1;
            arr1[c + 1 - j] = vec2[y - 1];
            for (u = vec2.begin(); u != vec2.end(); u++) {
              if (*u == arr1[c + 1 - j]) {
                vec2.erase(u);
                break;
              }
            }
          }
          if (n > 15) {
            l = l - (n - 15);
            r = r - (n - 15);
          }
          for (int j = l; j <= r; j++) {
            ans = ans + arr1[j];
          }
          cout << ans << "\n";
        }
      } else {
        int b;
        cin >> b;
        m = m + b;
      }
    }
  }
}