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;
namespace IO {
const int MAXSIZE = 1 << 20;
char buf[MAXSIZE], *p1, *p2;
inline int rd() {
  int x = 0;
  char c =
      (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, MAXSIZE, stdin), p1 == p2)
           ? EOF
           : *p1++);
  while (!isdigit(c)) {
    c = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, MAXSIZE, stdin), p1 == p2)
             ? EOF
             : *p1++);
  }
  while (isdigit(c))
    x = x * 10 + (c ^ 48),
    c = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, MAXSIZE, stdin), p1 == p2)
             ? EOF
             : *p1++);
  return x;
}
}  // namespace IO
using namespace IO;
int i, j, k, t, a[20], sb;
long long jc[20] = {1}, cur = 0, l, r, res, n;
void get(long long cur, int n) {
  int i, j, k, tmp;
  bool used[20] = {0};
  for (i = 1; i <= n; i++) {
    tmp = 1;
    while (cur >= jc[n - i]) {
      cur -= jc[n - i];
      tmp++;
    }
    for (j = 1, k = 0; j <= n; j++) {
      if (!used[j]) {
        k++;
      }
      if (k == tmp) {
        used[j] = 1;
        a[i] = j;
        break;
      }
    }
  }
}
int main() {
  for (i = 1; i <= 16; i++) {
    jc[i] = jc[i - 1] * i;
  }
  n = rd();
  t = rd();
  get(0, min(16, (int)n));
  while (t--) {
    k = rd();
    if (k == 2) {
      k = rd();
      cur += k;
      sb = 1;
    } else {
      l = rd();
      r = rd();
      if (n <= 16) {
        if (sb) {
          get(cur, n);
          sb = 0;
        }
        for (i = l, res = 0; i <= r; i++) {
          res += a[i];
        }
        printf("%lld\n", res);
        continue;
      }
      if (r <= n - 16) {
        printf("%lld\n", (r * r + r - l * l + l) / 2);
      } else {
        if (sb) {
          get(cur, 16);
          sb = 0;
        }
        if (l <= n - 16) {
          for (i = 1, res = 0; i <= r - (n - 15) + 1; i++) {
            res += n - 16 + a[i];
          }
          printf("%lld\n",
                 ((n - 16) * (n - 16) + (n - 16) - l * l + l) / 2 + res);
        } else {
          for (i = l - (n - 15) + 1, res = 0; i <= r - (n - 15) + 1; i++) {
            res += n - 16 + a[i];
          }
          printf("%lld\n", res);
        }
      }
    }
  }
}