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 fact[16];
long long topperm = 0;
vector<int> orgperm;
vector<int> perm;
void xthperm(long long x) {
  vector<int> temp = orgperm;
  topperm += x;
  x = topperm;
  perm.clear();
  int curfact = temp.size() - 1;
  while (curfact >= 0) {
    int ind = x / fact[curfact];
    x %= fact[curfact];
    curfact--;
    perm.push_back(temp[ind]);
    temp.erase(temp.begin() + ind);
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  fact[0] = 1;
  for (int i = 1; i <= 15; i++) fact[i] = i * fact[i - 1];
  int n, q;
  cin >> n >> q;
  if (n > 15) {
    for (int i = n - 15 + 1; i <= n; i++) orgperm.push_back(i);
    perm = orgperm;
    while (q--) {
      int qq;
      cin >> qq;
      if (qq == 1) {
        long long l, r;
        cin >> l >> r;
        long long sum = 0;
        if (l < n - 15 + 1) {
          long long sag;
          if (r >= n - 15 + 1)
            sag = n - 15;
          else
            sag = r;
          sum += (l + sag) * (sag - l + 1) / 2;
        }
        if (r >= n - 15 + 1) {
          long long sol;
          if (l < n - 15 + 1)
            sol = n - 15 + 1;
          else
            sol = l;
          for (int i = sol; i <= r; i++) {
            sum += perm[i - (n - 15 + 1)];
          }
        }
        cout << sum << '\n';
      } else {
        long long x;
        cin >> x;
        xthperm(x);
      }
    }
  } else {
    for (int i = 1; i <= n; i++) orgperm.push_back(i);
    perm = orgperm;
    while (q--) {
      int qq;
      cin >> qq;
      if (qq == 1) {
        int l, r;
        cin >> l >> r;
        int sum = 0;
        for (int i = l; i <= r; i++) sum += perm[i - 1];
        cout << sum << '\n';
      } else {
        long long x;
        cin >> x;
        xthperm(x);
      }
    }
  }
}