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>
static const int MOD = 1000000007;
using ll = long long;
using u32 = unsigned;
using u64 = unsigned long long;
using namespace std;
template <class T>
constexpr T INF = ::numeric_limits<T>::max() / 32 * 15 + 208;
int main() {
  int n, q;
  cin >> n >> q;
  int m = min(14, n);
  vector<ll> f(m + 1, 1);
  for (int i = 1; i <= m; ++i) {
    f[i] = f[i - 1] * i;
  }
  vector<ll> S(n + 1);
  for (int i = 0; i < n; ++i) S[i + 1] = S[i] + i + 1;
  vector<bool> used(m);
  auto set_p = [&](ll k) {
    fill(used.begin(), used.end(), 0);
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < m; ++j) {
        if (used[j]) continue;
        if (f[m - i - 1] <= k) {
          k -= f[m - i - 1];
          continue;
        }
        used[j] = 1;
        S[n - m + i + 1] = S[n - m + i] + n - m + j + 1;
        break;
      }
    }
  };
  ll val = 0;
  for (int i = 0; i < q; ++i) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int l, r;
      scanf("%d %d", &l, &r);
      printf("%lld\n", S[r] - S[l - 1]);
    } else {
      int k;
      scanf("%d", &k);
      val += k;
      set_p(val);
    }
  }
  return 0;
}