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;
vector<long long> ithPermutation(int n, long long i) {
  long long j, k = 0;
  vector<long long> fact(n), perm(n);
  fact[k] = 1;
  while (++k < n) fact[k] = fact[k - 1] * k;
  for (k = 0; k < n; ++k) {
    perm[k] = i / fact[n - 1 - k];
    i = i % fact[n - 1 - k];
  }
  for (k = n - 1; k > 0; --k)
    for (j = k - 1; j >= 0; --j)
      if (perm[j] <= perm[k]) perm[k]++;
  return perm;
}
int main() {
  int n, q;
  cin >> n >> q;
  long long s = 0;
  for (int i = 0; i < q; i++) {
    int a;
    cin >> a;
    if (a == 1) {
      long long l, r;
      cin >> l >> r;
      r--;
      l--;
      long long x = 0;
      if (n <= 15) {
        vector<long long> u = ithPermutation(n, s);
        for (int j = l; j <= r; j++) {
          x += u[j] + 1;
        }
      } else {
        if (l < n - 15) {
          long long r1 = min(r, (long long)(n - 16));
          x = ((r1 - l + 1) * (r1 + l + 2)) / 2;
          vector<long long> u = ithPermutation(15, s);
          for (int j = n - 15; j <= r; j++) {
            x += u[j - n + 15] + n - 14;
          }
        } else {
          vector<long long> u = ithPermutation(15, s);
          for (int j = l; j <= r; j++) {
            x += u[j - n + 15] + n - 14;
          }
        }
      }
      cout << x << endl;
    } else {
      long long b;
      cin >> b;
      s += b;
    }
  }
}