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;
constexpr int max_len = 14;
vector<long long> factorial;
vector<int> nth_permutation(long long n, int len) {
  vector<int> v(min(len, max_len));
  list<int> possible_num;
  for (int i = len - v.size() + 1; i <= len; i++) possible_num.push_back(i);
  --n;
  for (int i = 0; i < v.size(); i++) {
    auto it = possible_num.begin();
    advance(it, n / factorial[v.size() - i - 1]);
    v[i] = *it;
    possible_num.erase(it);
    n %= factorial[v.size() - i - 1];
  }
  return v;
}
void test_case() {
  int n, q;
  cin >> n >> q;
  long long x = 1;
  vector<int> v(min(n, max_len));
  vector<long long> prefix(v.size());
  bool need_update = true;
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int l, r;
      cin >> l >> r;
      if (r <= n - v.size())
        cout << (long long)(l + r) * (r - l + 1) / 2 << '\n';
      else {
        if (need_update) {
          v = nth_permutation(x, n);
          prefix[0] = v[0];
          for (int i = 1; i < v.size(); i++) prefix[i] = prefix[i - 1] + v[i];
          need_update = false;
        }
        long long ans = prefix[r - n + v.size() - 1];
        if (l <= n - v.size() + 1)
          ans += (long long)(l + n - v.size()) * (n - v.size() - l + 1) / 2;
        else
          ans -= prefix[l - n + v.size() - 2];
        cout << ans << '\n';
      }
    } else {
      long long tmp;
      cin >> tmp;
      x += tmp;
      need_update = true;
    }
  }
}
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  factorial.resize(max_len);
  factorial[0] = 1;
  for (int i = 1; i < max_len; i++) factorial[i] = factorial[i - 1] * i;
  int Case = 1;
  for (int i = 1; i <= Case; i++) {
    test_case();
  }
  return 0;
}