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 A, typename B>
string to_string(pair<A, B> p);
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);
string to_string(const string &s) { return '"' + s + '"'; }
string to_string(const char *s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
string to_string(vector<bool> v) {
  bool first = true;
  string res = "{";
  for (int i = 0; i < static_cast<int>(v.size()); i++) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(v[i]);
  }
  res += "}";
  return res;
}
template <size_t N>
string to_string(bitset<N> v) {
  string res = "";
  for (size_t i = 0; i < N; i++) {
    res += static_cast<char>('0' + v[i]);
  }
  return res;
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto &x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
         to_string(get<2>(p)) + ")";
}
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + "," +
         to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";
}
void debug() { cerr << endl; }
template <typename Head, typename... Tail>
void debug(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug(T...);
}
const int INF = 0x3f3f3f3f;
const int maxn = 30000 + 5;
static const long long ans[] = {1,
                                1,
                                2,
                                6,
                                24,
                                120,
                                720,
                                5040,
                                40320,
                                362880,
                                3628800LL,
                                39916800LL,
                                479001600LL,
                                6227020800LL,
                                87178291200LL};
vector<int> y;
void ni_Cantor(int len, long long num) {
  y.clear();
  vector<int> a;
  if (len <= 14) {
    for (int i = 1; i <= len; i++) a.push_back(i);
  } else {
    for (int i = 1; i <= 14; i++) a.push_back(len - 14 + i);
  }
  if (len > 14) len = 14;
  for (int i = 1; i <= len; i++) {
    int t = num / ans[len - i];
    num = num % ans[len - i];
    y.push_back(a[t]);
    a.erase(a.begin() + t);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  long long num = 0;
  int n, q;
  cin >> n >> q;
  ni_Cantor(n, 0);
  while (q--) {
    int f, f1, f2;
    cin >> f;
    if (f == 1) {
      cin >> f1 >> f2;
      if (n <= 14) {
        long long ans = 0;
        for (int i = f1 - 1; i < f2; i++) ans += y[i];
        cout << ans << endl;
      } else {
        long long ans = 0;
        for (int i = max(n - 14 + 1, f1); i <= f2; i++) {
          ans += y[i - (n - 14 + 1)];
        }
        long long pos = min(f2, n - 14);
        if (pos >= f1) {
          ans += (long long)(pos + f1) * (long long)(pos - f1 + 1) / 2;
        }
        cout << ans << endl;
      }
    } else {
      cin >> f2;
      num += f2;
      ni_Cantor(n, num);
    }
  }
  return 0;
}