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 mod = 1e9 + 7;
const long long N = 1e7 + 10;
long long findFirstNumIndex(long long& k, long long n) {
  if (n == 1) return 0;
  n--;
  long long first_num_index;
  long long n_partial_fact = n;
  while (k >= n_partial_fact && n > 1) {
    n_partial_fact = n_partial_fact * (n - 1);
    n--;
  }
  first_num_index = k / n_partial_fact;
  k = k % n_partial_fact;
  return first_num_index;
}
vector<long long> findKthPermutation(long long n, long long k) {
  vector<long long> ans;
  set<long long> s;
  for (long long i = 1; i <= n; i++) s.insert(i);
  set<long long>::iterator itr;
  itr = s.begin();
  k = k - 1;
  for (long long i = 0; i < n; i++) {
    long long index = findFirstNumIndex(k, n - i);
    advance(itr, index);
    ans.push_back(*itr);
    s.erase(itr);
    itr = s.begin();
  }
  return ans;
}
int32_t main() {
  long long n, q;
  cin >> n >> q;
  long long curr = 1;
  vector<long long> pref(n + 1);
  for (long long i = 1; i <= n; i++) {
    pref[i] = pref[i - 1] + i;
  }
  while (q--) {
    long long type;
    cin >> type;
    if (type == 1) {
      long long l, r;
      cin >> l >> r;
      if (n <= 15) {
        vector<long long> temp = findKthPermutation(n, curr);
        l--;
        r--;
        long long ans = 0;
        for (long long i = l; i <= r; i++) {
          ans += temp[i];
        }
        cout << ans << endl;
      } else {
        vector<long long> temp = findKthPermutation(15, curr);
        long long end = n - 15;
        for (long long j = 0; j < 15; j++) {
          temp[j] += end;
        }
        long long ans = 0;
        if (r <= end) {
          ans = pref[r] - pref[l - 1];
        } else if (l > end) {
          for (long long j = l; j <= r; j++) {
            ans += temp[j - end - 1];
          }
        } else {
          ans += pref[end] - pref[l - 1];
          for (long long i = end + 1; i <= r; i++) {
            ans += temp[i - end - 1];
          }
        }
        cout << ans << endl;
      }
    } else {
      long long x;
      cin >> x;
      curr += x;
    }
  }
}