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;
const long long mod = 1e9 + 7;
const long long MAX = 2e5 + 1;
void Solve() {
  long long n, q, p = 18;
  cin >> n >> q;
  vector<long long> a(n), prefix(n);
  iota(a.begin(), a.end(), 1);
  for (long long i = 0; i < n; i++) {
    prefix[i] = (i ? prefix[i - 1] + a[i] : a[i]);
  }
  function<void(vector<long long>)> fixPrefix =
      [&](vector<long long> permutation) {
        long long x = permutation.size();
        for (long long i = 0; i < x; i++) {
          a[n - x + i] = permutation[i];
        }
        for (long long i = n - x; i < n; i++) {
          prefix[i] = (i ? prefix[i - 1] + a[i] : a[i]);
        }
      };
  function<void(long long, long long, long long)> kThPermutation =
      [&](long long n, long long k, long long start) {
        vector<long long> fact(p + 1, 1);
        for (long long i = 1; i <= p; i++) {
          fact[i] = i * fact[i - 1];
        }
        vector<long long> permutation(n, start);
        for (long long i = 0; i < n; i++) {
          permutation[i] += k / fact[n - 1 - i];
          k %= fact[n - 1 - i];
        }
        for (long long i = n - 1; i >= 0; i--) {
          for (long long j = i - 1; j >= 0; j--) {
            if (permutation[j] <= permutation[i]) {
              permutation[i]++;
            }
          }
        }
        fixPrefix(permutation);
      };
  long long k = 0;
  while (q--) {
    long long type;
    cin >> type;
    if (type == 1) {
      long long l, r;
      cin >> l >> r;
      l--, r--;
      cout << prefix[r] - (l ? prefix[l - 1] : 0) << "\n";
    } else {
      long long x;
      cin >> x;
      k += x;
      kThPermutation(min(n, p), k, max(n - p + 1, 1LL));
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long testcases = 1;
  while (testcases--) {
    Solve();
  }
}