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 <typename T>
ostream &operator<<(ostream &out, const vector<T> &V);
template <typename T>
ostream &operator<<(ostream &out, const set<T> &S);
template <typename T1, typename T2>
ostream &operator<<(ostream &out, const pair<T1, T2> &P);
template <typename T1, typename T2>
ostream &operator<<(ostream &out, const map<T1, T2> &M);
int find_first_num_index(long long &k, long long n) {
  if (n == 1) {
    return 0;
  }
  n--;
  int first_num_index;
  long long n_partial_fact = n;
  while (k >= n_partial_fact && n > 1) {
    n_partial_fact *= (n - 1);
    n--;
  }
  first_num_index = k / n_partial_fact;
  k %= n_partial_fact;
  return first_num_index;
}
vector<int> find_kth_permutation(long long n, long long k) {
  vector<int> Ans;
  set<int> S;
  for (int i = 1; i <= n; i++) {
    S.insert(i);
  }
  auto itr = S.begin();
  k -= 1;
  for (int i = 0; i < n; i++) {
    int index = find_first_num_index(k, n - i);
    advance(itr, index);
    Ans.push_back((*itr));
    S.erase(itr);
    itr = S.begin();
  }
  return Ans;
}
void testcase() {
  int n, q;
  cin >> n >> q;
  int suff_len = min(14, n);
  int pref_len = n - suff_len;
  long long x = 1;
  long long Sil[20];
  Sil[0] = 1;
  vector<int> Suf = find_kth_permutation(suff_len, 1);
  for (int i = 1; i <= suff_len; i++) {
    Sil[i] = Sil[i - 1] * i;
  }
  for (int h = 0; h < q; h++) {
    int t;
    cin >> t;
    if (t == 1) {
      int l, r;
      cin >> l >> r;
      long long ans = 0;
      int r_pr = min(r, pref_len);
      if (l <= r_pr) {
        ans += 1ll * (l + r_pr) * (r_pr - l + 1) / 2;
      }
      r -= pref_len + 1;
      l -= pref_len + 1;
      r_pr = max(r_pr - pref_len - 1, 0);
      for (int i = max(l, 0); i <= r; i++) {
        ans += pref_len + Suf[i];
      }
      cout << ans << endl;
    } else {
      int dx;
      cin >> dx;
      x += dx;
      x = min(x, Sil[suff_len]);
      Suf = find_kth_permutation(suff_len, x);
    }
  }
}
void init() {}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  init();
  int T = 1;
  for (int t = 1; t <= T; t++) {
    testcase();
  }
  return 0;
}
template <typename T>
ostream &operator<<(ostream &out, const vector<T> &V) {
  if (!V.empty()) {
    out << "{";
    for (auto v : V) out << v << ", ";
    out << "\b\b}";
  } else {
    out << "{}";
  }
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, const set<T> &S) {
  if (!S.empty()) {
    out << "set{";
    for (auto v : S) out << v << ", ";
    out << "\b\b}";
  } else {
    out << "{}";
  }
  return out;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &out, const map<T1, T2> &M) {
  if (!M.empty()) {
    out << "map{";
    for (auto &kv : M)
      out << "(" << kv.first << ": " << kv.second << ")"
          << ", ";
    out << "\b\b}";
  } else {
    out << "{}";
  }
  return out;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &out, const pair<T1, T2> &P) {
  out << "{" << P.first << ", " << P.second << "}";
  return out;
}