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<long long> genperm(long long x, set<long long> p,
                          vector<long long> &fac) {
  int n = p.size();
  vector<long long> ans;
  for (int i = 0; i < n && x > 0; ++i) {
    if (x >= fac[n - i - 1]) {
      long long need = (x + fac[n - i - 1] - 1) / fac[n - i - 1];
      int j = 0;
      auto it = p.begin();
      while (j < need) ++j, ++it;
      it--, j--;
      ans.push_back(*it);
      p.erase(it);
      x -= fac[n - i - 1] * j;
    } else {
      ans.push_back(*p.begin());
      p.erase(p.begin());
    }
  }
  while (!p.empty()) {
    ans.push_back(*p.begin());
    p.erase(p.begin());
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, q;
  cin >> n >> q;
  vector<long long> fac(16, 1);
  for (long long i = 1; i <= 15; ++i) fac[i] = fac[i - 1] * i;
  long long x = 0;
  set<long long> p;
  for (long long i = n; i && i > n - 15; --i) p.insert(i);
  vector<long long> curperm = genperm(1, p, fac);
  while (q--) {
    int qi;
    cin >> qi;
    if (qi == 2) {
      long long xi;
      cin >> xi;
      x += xi;
      curperm = genperm(x + 1, p, fac);
    } else {
      long long l, r, sum = 0;
      cin >> l >> r;
      if (n <= 15)
        for (int i = l - 1; i < r; ++i) sum += curperm[i];
      else {
        long long mn = min(r, (long long)n - 15);
        if (l <= n - 15) {
          sum += (long long)(mn * (mn + 1) / 2) - (long long)(l * (l - 1) / 2);
          if (r > n - 15)
            for (int i = 0, j = n - 14; j <= r; ++j, ++i) sum += curperm[i];
        } else
          for (int i = l - (n - 14); i <= r - (n - 14); ++i) sum += curperm[i];
      }
      cout << sum << endl;
    }
  }
  return 0;
}