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;
using PII = pair<int, int>;
using VI = vector<int>;
ll pref_sum(ll n) { return (n * (n + 1)) / 2; }
vector<int> gen(int sz, ll id) {
  VI ranks(sz);
  ll div = 1;
  for (int i = 1; i <= sz; i++) {
    div *= i;
  }
  for (int i = 1; i <= sz; i++) {
    div /= (sz - i + 1);
    ranks[i - 1] = id / div;
    id %= div;
  }
  set<int> vals;
  for (int i = 1; i <= sz; i++) {
    vals.insert(i);
  }
  VI perm;
  for (int r : ranks) {
    auto it = vals.begin();
    while (r > 0) {
      it++;
      r--;
    }
    perm.push_back(*it);
    vals.erase(it);
  }
  return perm;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, q;
  cin >> n >> q;
  int m = min(n, 15);
  ll id = 0;
  while (q--) {
    int ty;
    cin >> ty;
    if (ty == 1) {
      int l, r;
      cin >> l >> r;
      ll ans = 0;
      if (l <= n - m) {
        ll r1 = min(r, n - m);
        ans += pref_sum(r1) - pref_sum(l - 1);
      }
      if (r > n - m) {
        auto suf = gen(m, id);
        for (int i = max(l, n - m + 1); i <= r; i++) {
          ans += n - m + suf[i - (n - m) - 1];
        }
      }
      cout << ans << endl;
    } else {
      int x;
      cin >> x;
      id += x;
    }
  }
  return 0;
}