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;
const long long N = 2e5 + 5;
long long a[N];
long long s[N];
long long gt[20];
long long n, q, m = 15;
void sub1() {
  long long cur = 1;
  while (q--) {
    long long typ;
    cin >> typ;
    if (typ == 1) {
      long long l, r;
      cin >> l >> r;
      vector<bool> flag(n + 1, 0);
      for (long long i = 1, sum = 0; i <= n; i++) {
        for (long long val = 1; val <= n; val++)
          if (!flag[val]) {
            if (sum + gt[n - i] < cur) {
              sum += gt[n - i];
            } else {
              a[i] = val;
              flag[val] = true;
              break;
            }
          }
      }
      long long sum = 0;
      for (long long i = l; i <= r; i++) {
        sum += a[i];
      }
      cout << sum << "\n";
    } else {
      long long x;
      cin >> x;
      cur += x;
    }
  }
}
void sub2() {
  vector<long long> val(0);
  long long lim = n - m;
  for (long long i = lim + 1; i <= n; i++) {
    val.push_back(i);
  }
  for (long long i = 1; i <= n - m + 1; i++) {
    s[i] = s[i - 1] + i;
  }
  long long cur = 1;
  while (q--) {
    long long typ;
    cin >> typ;
    if (typ == 1) {
      long long l, r;
      cin >> l >> r;
      vector<bool> flag(m, 0);
      for (long long i = 1, sum = 0; i <= m; i++) {
        for (long long j = 0; j < m; j++)
          if (!flag[j]) {
            if (sum + gt[m - i] < cur) {
              sum += gt[m - i];
            } else {
              a[lim + i] = val[j];
              flag[j] = true;
              break;
            }
          }
      }
      long long sum = 0;
      if (l <= lim) sum += s[lim] - s[l - 1];
      if (r <= lim) sum -= s[lim] - s[r];
      for (long long i = max(lim + 1, l); i <= r; i++) {
        sum += a[i];
      }
      cout << sum << "\n";
    } else {
      long long x;
      cin >> x;
      cur += x;
    }
  }
}
void proc() {
  gt[0] = 1;
  for (long long i = 1; i <= 15; i++) {
    gt[i] = gt[i - 1] * i;
  }
  if (n <= 15)
    sub1();
  else
    sub2();
}
void enter() { cin >> n >> q; }
void preproc() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
signed main() {
  preproc();
  enter();
  proc();
  return 0;
}