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;
vector<long long> fct(18);
long long konumbul(long long len, long long val, long long times) {
  long long check = min(len, 15LL);
  set<long long> s;
  for (long long i = 0; i < check; i++) {
    s.insert(len - i);
  }
  for (long long i = 0; i < check; i++) {
    long long change = fct[check - 1 - i];
    long long bul = times / change;
    set<long long>::iterator it = s.begin();
    for (long long j = 0; j < bul; j++) {
      it++;
    }
    if (check - i == val) return *it;
    s.erase(it);
    times -= bul * change;
  }
}
void solve() {
  long long n, q, perm = 0;
  fct[0] = 1;
  for (long long i = 1; i < fct.size(); i++) {
    fct[i] = fct[i - 1] * i;
  }
  cin >> n >> q;
  vector<long long> sum(n + 1, 0);
  vector<long long> v(n);
  for (long long i = 1; i <= n; i++) v[i - 1] = i;
  for (long long i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + i;
  }
  for (long long i = 0; i < q; i++) {
    long long x, a, b;
    cin >> x;
    if (x == 1) {
      cin >> a >> b;
      long long res = sum[b] - sum[a - 1];
      long long check = min(n, 15LL);
      for (long long i = 0; i < check; i++) {
        long long needed = konumbul(n, i + 1, perm);
        if (n - i < a) break;
        if (n - i <= b) res += needed - (n - i);
      }
      cout << res << endl;
    } else {
      cin >> a;
      perm += a;
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long T = 1;
  while (T--) solve();
}