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;
using ll = long long;
const vector<ll> fac = {1,       1,        2,         6,         24,
                        120,     720,      5040,      40320,     362880,
                        3628800, 39916800, 479001600, 6227020800};
vector<int> kth(vector<int>& R, ll k) {
  vector<int> num = R, ret;
  for (int i = R.size() - 1; ~i; --i) {
    ll x = k / fac[i];
    ret.emplace_back(num[x]);
    num.erase(num.begin() + x);
    k %= fac[i];
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  int num_right = min(n, 14);
  int num_left = n - num_right;
  vector<ll> psum(num_left + 1);
  vector<int> R(num_right);
  iota(psum.begin() + 1, psum.end(), 1);
  iota(R.begin(), R.end(), num_left + 1);
  partial_sum(psum.begin(), psum.end(), psum.begin());
  ll k = 0;
  for (int op, x, l, r; q--;) {
    cin >> op;
    if (op == 1) {
      cin >> l >> r;
      ll ans = 0;
      if (l <= num_left) ans += psum[min(num_left, r)] - psum[l - 1];
      if (r > num_left) {
        auto vt = kth(R, k);
        for (int i = max(0, l - num_left - 1); i < r - num_left; ++i)
          ans += vt[i];
      }
      cout << ans << '\n';
    } else {
      cin >> x;
      k += x;
    }
  }
}