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;
ifstream fin("Sisend.txt");
long long n, q, aa, a, aaa;
bool b[20];
vector<long long> v, vv, kor;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);
  cin >> n >> q;
  long long nn = n;
  long long p = 1;
  for (int(i) = (0); ((i)) < ((15)); ((i))++) {
    kor.emplace_back(p);
    vv.emplace_back(nn);
    p *= (i + 2);
    nn--;
    if (!nn) break;
  }
  reverse(vv.begin(), vv.end());
  reverse(kor.begin(), kor.end());
  v = vv;
  long long su = 0;
  su = max(su, vv[0] - 1);
  while (q--) {
    long long l, r, c, vas = 0;
    cin >> c;
    if (c == 1) {
      cin >> l >> r;
      if (l <= su) {
        if (r <= su) {
          cout << (l + r) * (r - l + 1) / 2 << endl;
          continue;
        } else {
          vas += (l + su) * (su - l + 1) / 2;
        }
      }
      l = max((long long)0, l - su - 1);
      r = r - su - 1;
      if (r > -1)
        for (long long i = l; i <= r; i++) vas += v[i];
      cout << vas << endl;
    } else {
      cin >> aa;
      a += aa;
      aaa = a;
      long long in = 0;
      memset(b, 0, sizeof(b));
      v.resize(0);
      while (in < vv.size() - 1) {
        long long m = aaa / kor[in + 1];
        aaa %= kor[in + 1];
        for (int(j) = (0); ((j)) < ((vv.size())); ((j))++)
          if (!b[j])
            if (m)
              m--;
            else {
              v.emplace_back(vv[j]);
              b[j] = 1;
              break;
            }
        in++;
      }
      for (int(j) = (0); ((j)) < ((vv.size())); ((j))++)
        if (!b[j]) v.emplace_back(vv[j]);
    }
  }
}