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, q;
void solve() {
  long long sum = 0;
  vector<long long> po(20);
  po[1] = 1;
  for (long long i = 2; i < 20; i++) {
    po[i] = po[i - 1] * i;
  }
  while (q--) {
    long long t;
    cin >> t;
    if (t == 2) {
      long long x;
      cin >> x;
      sum += x;
    } else {
      long long l, r;
      cin >> l >> r;
      long long need = sum;
      set<long long> st;
      vector<long long> u;
      for (long long i = 1; i <= n; i++) st.insert(i);
      for (long long i = 0; i < n; i++) {
        if (need == 0) {
          u.push_back(*st.begin());
          st.erase(st.begin());
        } else {
          long long val = 0;
          for (auto to : st) {
            if (need < po[n - i - 1]) {
              val = to;
              break;
            } else {
              need -= po[n - i - 1];
            }
          }
          u.push_back(val);
          st.erase(val);
        }
      }
      long long ans = 0;
      for (long long i = l - 1; i < r; i++) {
        ans += u[i];
      }
      cout << ans << endl;
    }
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> q;
  if (n < 20) {
    solve();
  } else {
    vector<long long> po(20, 1);
    for (long long i = 1; i < 20; i++) {
      po[i] = po[i - 1] * i;
    }
    long long need = 0, sum = 0;
    while (q--) {
      long long t;
      cin >> t;
      if (t == 2) {
        long long x;
        cin >> x;
        sum += x;
      } else {
        long long l, r;
        cin >> l >> r;
        long long pos = min(n - 20, r), ans = 0;
        if (l <= pos) {
          ans += pos * (pos + 1) / 2 - l * (l - 1) / 2;
        }
        set<long long> st;
        for (long long i = n - 20 + 1; i <= n; i++) st.insert(i);
        long long need = sum;
        for (long long i = n - 20; i < n; i++) {
          long long kek = 0;
          if (need == 0) {
            kek = *st.begin();
            st.erase(kek);
          } else {
            for (auto to : st) {
              if (need < po[n - i - 1]) {
                kek = to;
                break;
              } else {
                need -= po[n - i - 1];
              }
            }
            st.erase(kek);
          }
          if (l <= i + 1 && i + 1 <= r) {
            ans += kek;
          }
        }
        cout << ans << endl;
      }
    }
  }
  return 0;
}