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 dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
const int dx8[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dy8[] = {0, -1, 0, 1, 1, -1, 1, -1};
vector<long long> factoradic_perm(int n, long long step) {
  vector<long long> ret, cand;
  vector<long long> fact(1, 1);
  int i;
  for (int i = 0; i < n; i++) {
    cand.push_back(i);
    fact.push_back(fact.back() * (i + 1));
  }
  for (int i = 0; i < n; i++) {
    int num = step / fact[n - 1 - i];
    step %= fact[n - 1 - i];
    ret.push_back(cand[num]);
    cand.erase(cand.begin() + num);
  }
  return ret;
}
long long sum(long long x) { return (x * (x + 1)) / 2; }
long long max(long long a, long long b) {
  if (a < b) return b;
  return a;
}
void solve() {
  long long n, q;
  cin >> n;
  long long k = 0;
  cin >> q;
  while (q--) {
    long long a;
    cin >> a;
    if (a == 1) {
      long long l, r;
      cin >> l >> r;
      long long x0 = max(0, n - 14);
      if (r <= x0) {
        cout << sum(r) - sum(l - 1) << '\n';
      } else if (l <= x0 && x0 <= r) {
        long long ans = sum(x0) - sum(l - 1);
        vector<long long> arr = factoradic_perm(n - x0, k);
        for (long long i = x0 + 1; i <= r; i++) {
          ans += arr[i - x0 - 1] + x0 + 1;
        }
        cout << ans << '\n';
      } else {
        long long ans = 0;
        vector<long long> arr = factoradic_perm(n - x0, k);
        for (long long i = l; i <= r; i++) {
          ans += arr[i - x0 - 1] + x0 + 1;
        }
        cout << ans << '\n';
      }
    } else {
      long long x;
      cin >> x;
      k += x;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
}