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 N = 2e5 + 10;
void permutation(int *a, int len, long long n) {
  long long pn = 1;
  set<int> st;
  int t;
  for (int i = 1; i < len; ++i) pn *= i;
  n %= (pn * len);
  for (int i = 1; i <= len; ++i) st.insert(i);
  for (int i = 0; i < len - 1; ++i) {
    t = n / pn, n %= pn, pn /= len - 1 - i;
    auto iter = st.begin();
    while (t--) ++iter;
    a[i] = *iter;
    st.erase(iter);
  }
  a[len - 1] = *st.begin();
}
long long psum[N];
int n, l, r, x, tp, q, a[N];
long long pn = 0;
int main() {
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) a[i] = i, psum[i] = psum[i - 1] + a[i];
  for (int i = 0; i < q; ++i) {
    cin >> tp;
    if (tp == 1) {
      if (n <= 15)
        permutation(a + 1, n, pn);
      else {
        permutation(a + 1 + n - 15, 15, pn);
        for (int i = 0; i < 15; ++i) {
          a[1 + n - 15 + i] += n - 15;
        }
      }
      long long sum = 0;
      cin >> l >> r;
      if (r >= n - 14) {
        for (int i = max(n - 14, l); i <= r; ++i) sum += a[i];
      }
      if (l <= n - 15) {
        sum += psum[min(n - 15, r)] - psum[l - 1];
      }
      cout << sum << endl;
    } else if (tp == 2) {
      cin >> x;
      pn += x;
    } else
      assert(0);
  }
  return 0;
}