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 MAX_SIZE = 15;
vector<long long> nums;
long long n, q;
long long current_id = 0;
long long fact[MAX_SIZE + 1];
long long sum(long long l, long long r) {
  long long res = 0;
  if (n <= MAX_SIZE) {
    for (int i = l; i <= r; ++i) {
      res += nums[i];
    }
  } else {
    if (l > n - 16) {
      for (int i = l; i <= r; ++i) {
        res += nums[i];
      }
    } else {
      if (r > n - 16) {
        for (int i = n - 15; i <= r; ++i) {
          res += nums[i];
        }
        res += (n - 15) * (n - 14) / 2 - l * (l + 1) / 2;
      } else {
        res += (r + 1) * (r + 2) / 2 - l * (l + 1) / 2;
      }
    }
  }
  return res;
}
vector<long long> shift_to(long long id, long long size) {
  vector<long long> res;
  set<long long> to_be_inserted;
  for (int i = 1; i <= size; ++i) to_be_inserted.insert(i);
  for (int i = 1; i <= size; ++i) {
    long long num = id / fact[size - i];
    auto head = to_be_inserted.begin();
    for (int j = 0; j < num; ++j) {
      head++;
    }
    res.push_back(*head);
    to_be_inserted.erase(*head);
    id -= num * fact[size - i];
  }
  return res;
}
void shift(long long x) {
  current_id += x;
  if (n <= MAX_SIZE) {
    nums = shift_to(current_id, n);
  } else {
    vector<long long> temp = shift_to(current_id, 15);
    for (int i = n - 15; i < n; ++i) {
      nums[i] = temp[i - (n - 15)] + (n - 15);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  fact[0] = 1;
  for (int i = 1; i <= 15; ++i) {
    fact[i] = fact[i - 1] * i;
  }
  cin >> n >> q;
  nums.resize(n);
  for (int i = 0; i < n; ++i) {
    nums[i] = i + 1;
  }
  while (q--) {
    int option;
    cin >> option;
    if (option == 1) {
      long long l, r;
      cin >> l >> r;
      l--;
      r--;
      cout << sum(l, r) << "\n";
    } else {
      long long x;
      cin >> x;
      shift(x);
    }
  }
}