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 fact[16];
deque<int> sira;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  clock_t st = clock();
  long long n;
  cin >> n;
  int q;
  cin >> q;
  int last[15];
  fact[0] = 1;
  for (int i = 1; i <= 15; i++) {
    fact[i] *= i;
  }
  for (int i = min(14LL, n - 1); i >= 0; i--) {
    last[min(n, 15LL) - i - 1] = n - i;
  }
  long long minn = min(n, 15LL);
  long long curperm = 0;
  while (q--) {
    int type;
    cin >> type;
    long long ans = 0;
    if (type == 1) {
      long long l, r;
      cin >> l >> r;
      if (l < n - 14) {
        if (r < n - 14) {
          ans = (r * (r + 1) / 2) - ((l - 1) * l / 2);
        } else {
          ans = ((n - 15) * (n - 14) / 2) - ((l - 1) * l / 2);
          for (int i = 0; i <= r - n + 14; i++) ans += last[i];
        }
      } else {
        for (int i = l - n + minn - 1; i <= r - n + minn - 1; i++)
          ans += last[i];
      }
      cout << ans << '\n';
    } else if (type == 2) {
      long long x;
      cin >> x;
      curperm += x;
      int pre = 0;
      set<int> s;
      for (int i = min(14LL, n - 1); i >= 0; i--) s.insert(n - i);
      long long tmp = curperm;
      for (int i = 1; i <= min(n, 15LL) && s.size(); i++) {
        int mod = tmp % i;
        tmp /= i;
        sira.push_front(mod);
      }
      for (int i = 0; i < min(n, 15LL); i++) {
        int cur = sira.front();
        sira.pop_front();
        auto it = s.begin();
        advance(it, cur);
        last[i] = *it;
        s.erase(it);
      }
    }
  }
  cerr << setprecision(10) << fixed;
  cerr << (double)(clock() - st) / CLOCKS_PER_SEC << '\n';
}