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 int N = 15;
int A[N];
long long sums[N + 1];
long long counters[N - 1];
long long offset = 0;
long long factorial[N + 1];
inline int counter(int i) { return (offset / factorial[i + 1]) % (i + 2); }
void do_recompute(int n) {
  bitset<N> bs(0);
  for (int j = 0; j < min(n, N) - 1; j++) {
    int i = 0;
    int c = counter(min(n, N) - 2 - j);
    while (c >= 0) {
      if (!bs.test(i)) {
        c--;
      }
      i++;
    }
    A[j] = i;
    sums[j + 1] = sums[j] + A[j];
    bs.set(i - 1);
  }
  int i = 0;
  while (bs.test(i)) i++;
  A[min(n, N) - 1] = i + 1;
  sums[min(n, N)] = sums[min(n, N) - 1] + A[min(n, N) - 1];
}
int main() {
  int n, q;
  bool recompute = true;
  scanf("%d %d", &n, &q);
  long long base_sum =
      n > N ? ((((long long)(n - N)) * ((long long)(n - N + 1))) / 2LL) : 0LL;
  factorial[0] = 1;
  for (int i = 1; i < N + 1; i++) {
    factorial[i] = factorial[i - 1] * i;
  }
  for (int iq = 0; iq < q; iq++) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int l, r;
      scanf("%d %d", &l, &r);
      if (recompute) {
        do_recompute(n);
        recompute = false;
      }
      long long sum_l;
      long long sum_r;
      if (n > N) {
        sum_l =
            l - 1 >= n - N
                ? base_sum + sums[l - 1 - (n - N)] +
                      (((long long)(l - 1 - (n - N))) * ((long long)(n - N)))
                : (((long long)(l - 1)) * ((long long)(l))) / 2LL;
        sum_r = r >= n - N
                    ? base_sum + sums[r - (n - N)] +
                          (((long long)(r - (n - N))) * ((long long)(n - N)))
                    : (((long long)(r)) * ((long long)(r + 1))) / 2LL;
      } else {
        sum_l = sums[l - 1];
        sum_r = sums[r];
      }
      printf("%lld\n", sum_r - sum_l);
    } else {
      long long x;
      scanf("%lld", &x);
      offset += x;
      recompute = true;
    }
  }
  return 0;
}