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;
template <typename... Ts>
void do_nothing(Ts &&...vals) {}
constexpr long long MOD = 1000000007;
constexpr int INF = 0x7f7f7f7f;
constexpr double EPS = 1e-8;
long long fastPow(long long num, long long exp) {
  if (exp == 0) return 1;
  if (exp == 1)
    return num;
  else {
    long long half = fastPow(num, exp / 2);
    if (exp % 2 == 1) {
      return half * half * num;
    } else {
      return half * half;
    }
  }
}
int main(int argn, char **argv) {
  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);
  int n, q;
  cin >> n >> q;
  vector<long long> prefix(n + 1);
  prefix[0] = 0;
  for (int(i) = (1); (i) <= (n); (i)++) {
    prefix[i] = prefix[i - 1] + i;
  }
  long long advance = 0;
  vector<long long> modNum(20, (long long)1);
  int maxChanged = 2;
  for (; maxChanged <= n; maxChanged++) {
    modNum[maxChanged] = modNum[maxChanged - 1] * (maxChanged);
    if (modNum[maxChanged] > 2e10) break;
  }
  for (int(i) = (0); (i) <= ((q)-1); (i)++) {
    int kind;
    cin >> kind;
    if (kind == 1) {
      int l, r;
      cin >> l >> r;
      vector<long long> num(maxChanged);
      vector<bool> used(maxChanged, false);
      for (int(i) = (0); (i) <= ((maxChanged)-1); (i)++) {
        num[i] = n + 1 + i - maxChanged;
      }
      for (int j = maxChanged; j >= 1; j--) {
        long long newOrder = (advance % modNum[j]) / modNum[j - 1];
        int crtIndex = -1;
        for (int(s) = (0); (s) <= ((maxChanged)-1); (s)++) {
          if (!used[s]) {
            crtIndex++;
            if (crtIndex == newOrder) {
              used[s] = true;
              prefix[n + 1 - j] = prefix[n - j] + num[s];
              break;
            }
          }
        }
      }
      cout << prefix[r] - prefix[l - 1] << endl;
    } else {
      int x;
      cin >> x;
      advance += x;
    }
  }
}