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;
template <class T>
ostream &operator<<(ostream &str, vector<T> &a) {
  for (auto &i : a) {
    str << i << " ";
  }
  return str;
}
template <class T>
istream &operator>>(istream &str, vector<T> &a) {
  for (auto &i : a) {
    str >> i;
  }
  return str;
}
template <class T>
ostream &operator<<(ostream &str, pair<T, T> &a) {
  str << a.first << " " << a.second;
  return str;
}
template <class T>
istream &operator>>(istream &str, pair<T, T> &a) {
  str >> a.first >> a.second;
  return str;
}
void solve();
signed main() {
  srand(time(0));
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.setf(ios::fixed);
  cout.precision(20);
  solve();
  return 0;
}
const long long MAXN = 1e6 + 10, INF = 1e18 + 10;
vector<long long> factorial;
long long N = 15;
vector<long long> gen(long long first) {
  vector<long long> a(N);
  for (long long i = 0; i < N; ++i) {
    a[i] = i;
  }
  set<long long> se;
  for (auto it : a) {
    se.insert(it);
  }
  long long n = a.size();
  for (long long i = 0; i < n; ++i) {
    for (auto el : se) {
      if (first - factorial[n - i - 1] > 0) {
        first -= factorial[n - i - 1];
      } else {
        a[i] = el;
        se.erase(el);
        break;
      }
    }
  }
  return a;
}
void solve() {
  long long n, q;
  cin >> n >> q;
  vector<long long> ps(n + 1);
  vector<long long> a(n + 1);
  for (long long i = 1; i <= n; ++i) {
    a[i] = i;
  }
  ps[1] = 1;
  for (long long i = 2; i <= n; ++i) {
    ps[i] = ps[i - 1] + i;
  }
  N = min(n, N);
  factorial.resize(21);
  factorial[0] = 1;
  for (long long i = 1; i <= 20; ++i) {
    factorial[i] = factorial[i - 1] * i;
  }
  long long add = n - N + 1;
  vector<long long> dp(n + 1);
  long long sum = 0;
  for (long long _ = 0; _ < q; ++_) {
    long long type;
    cin >> type;
    if (type == 1) {
      long long l, r;
      cin >> l >> r;
      cout << ps[r] - ps[l - 1] << '\n';
    } else {
      long long first;
      cin >> first;
      sum += first;
      vector<long long> p = gen(sum + 1);
      for (auto &it : p) {
        it += add;
      }
      for (long long i = n - N + 1; i <= n; ++i) {
        a[i] = p[i - (n - N + 1)];
      }
      for (long long i = n - N + 1; i <= n; ++i) {
        ps[i] = ps[i - 1] + a[i];
      }
    }
  }
}