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 FAC[20];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  FAC[0] = 1;
  for (long long i = 1; i <= 15; i++) FAC[i] = i * FAC[i - 1];
  long long n, q;
  cin >> n >> q;
  if (n > 15) {
    long long cnt = 0;
    for (long long i = 0; i < q; i++) {
      long long k, l, x;
      cin >> k;
      if (k == 1) {
        cin >> l >> x;
        long long p = cnt;
        vector<long long> v;
        vector<long long> a;
        for (long long i = 1; i <= 15; i++) v.push_back(i);
        for (long long i = 15; i >= 1; i--) {
          long long r = p % FAC[i - 1];
          long long t = p / FAC[i - 1];
          p = r;
          a.push_back(v[t]);
          v.erase(v.begin() + t);
        }
        long long ans = 0;
        if (l > n - 15) {
          for (long long i = l - n + 14; i < x - n + 15; i++) ans += a[i];
          ans += (n - 15) * (x - l + 1);
        } else if (x <= n - 15)
          ans = (l + x) * (x - l + 1) / 2;
        else {
          ans += (l + n - 15) * (n - 14 - l) / 2;
          for (long long i = 0; i < x - n + 15; i++) ans += (n - 15) + a[i];
        }
        cout << ans << endl;
      } else {
        cin >> x;
        cnt += x;
      }
    }
  } else {
    long long cnt = 0;
    for (long long i = 0; i < q; i++) {
      long long k, l, x;
      cin >> k;
      if (k == 1) {
        cin >> l >> x;
        long long p = cnt;
        vector<long long> v;
        vector<long long> a;
        for (long long i = 1; i <= n; i++) v.push_back(i);
        for (long long i = n; i >= 1; i--) {
          long long r = p % FAC[i - 1];
          long long t = p / FAC[i - 1];
          p = r;
          a.push_back(v[t]);
          v.erase(v.begin() + t);
        }
        long long ans = 0;
        for (long long i = l - 1; i < x; i++) ans += a[i];
        cout << ans << endl;
      } else {
        cin >> x;
        cnt += x;
      }
    }
  }
  return 0;
}