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 n, q, a[200010], pos[200010], b[200010], f[20];
void init() {
  f[0] = 1;
  for (long long i = 1; i < 15; ++i) f[i] = i * f[i - 1];
}
long long solve(long long x, long long l, long long r) {
  long long d = 1;
  while (f[d] < x) ++d;
  long long s = n - d + 1;
  if (r < s) return (l + r) * (r - l + 1) / 2;
  long long pre = 0;
  if (l < s) pre = (l + s - 1) * (s - l) / 2;
  long long re = x - 1;
  bool used[n - s + 2];
  memset(used, 0, sizeof(used));
  for (long long i = s; i <= n; ++i) {
    if (i > r) return pre;
    long long one = f[n - i];
    long long rank = re / one + 1;
    re %= one;
    long long cur = 0;
    for (long long j = 1; j <= n - s + 1; ++j) {
      if (!used[j]) ++cur;
      if (cur == rank) {
        used[j] = 1;
        if (i >= l) pre += s - 1 + j;
        break;
      }
    }
  }
  return pre;
}
signed main() {
  init();
  cin >> n >> q;
  long long cnt = 1;
  while (q--) {
    long long t;
    cin >> t;
    if (t == 1) {
      long long l, r;
      cin >> l >> r;
      cout << solve(cnt, l, r) << endl;
    } else {
      long long x;
      cin >> x;
      cnt += x;
    }
  }
  return 0;
}