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 N, P, Q;
long long memo[1 << 14];
long long pre[200005];
long long perm = 1;
long long dp(long long msk) {
  if (msk == (1 << P) - 1) return memo[msk] = 1;
  if (memo[msk]) return memo[msk];
  long long ans = 0;
  for (long long i = 0; i < P; i++)
    if (!(msk & (1 << i))) ans += dp(msk | (1 << i));
  return memo[msk] = ans;
}
void getPerm(long long msk, long long perm, vector<long long>& v) {
  if (msk == (1 << P) - 1) return;
  for (long long i = 0; i < P; i++) {
    if (!(msk & (1 << i))) {
      if (perm <= memo[msk | (1 << i)]) {
        v.push_back(i);
        getPerm(msk | (1 << i), perm, v);
        return;
      }
      perm -= memo[msk | (1 << i)];
    }
  }
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> N >> Q;
  for (long long i = 1; i <= N; i++) pre[i] = pre[i - 1] + i;
  P = min(14LL, N);
  dp(0);
  while (Q--) {
    long long op, l, r;
    cin >> op >> l;
    if (op == 2) {
      perm += l;
    } else {
      cin >> r;
      long long lim = N - P;
      if (r <= lim) {
        cout << pre[r] - pre[l - 1] << "\n";
      } else {
        long long ans = 0;
        if (l <= lim) ans += pre[lim] - pre[l - 1];
        vector<long long> v;
        getPerm(0, perm, v);
        for (long long i = max(l, lim + 1); i <= r; i++) {
          ans += v[i - lim - 1] + lim + 1;
        }
        cout << ans << "\n";
      }
    }
  }
  return 0;
}