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 MAXN = pow(10, 5) * 2;
int n, q;
int a[16];
long long fac[16];
int was[16] = {0};
long long k;
bool updated = false;
int add;
void factorial() {
  fac[0] = 1;
  fac[1] = 1;
  for (int i = 2; i <= 15; i++) fac[i] = fac[i - 1] * i;
}
void gen() {
  long long tk = k;
  int limit = min(14, n);
  for (int i = 1; i <= limit; i++) {
    long long awas = tk / fac[limit - i];
    tk %= fac[limit - i];
    for (int j = 1; j <= limit; j++) {
      if (was[j] == 0) {
        if (awas == 0) {
          a[i] = j + add;
          was[j] = 1;
          break;
        }
        awas--;
      }
    }
  }
  for (int i = 0; i < 15; i++) was[i] = 0;
  return;
}
void solve() {
  cin >> n >> q;
  k = 0;
  if (n <= 14)
    add = 0;
  else
    add = n - 14;
  factorial();
  while (q--) {
    int x;
    cin >> x;
    if (x == 1) {
      if (!updated) {
        gen();
        updated = true;
      }
      int l, r;
      cin >> l >> r;
      long long tr, tl;
      if (n <= 14) {
        tr = -1;
        tl = l;
      } else {
        tr = min(r, n - 14);
        r -= n - 14;
        tl = max(l - (n - 14), 1);
      }
      long long sum = 0;
      if (tr >= l) sum = (l + tr) * (tr - l + 1) / 2;
      for (int i = tl; i <= r; i++) {
        sum += a[i];
      }
      cout << sum << "\n";
    } else {
      cin >> x;
      k += x;
      updated = false;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  solve();
  ;
  return 0;
}