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 gcd(long long x, long long y) { return (!y ? x : gcd(y, x % y)); }
const long long M = 15, maxn = 2e5 + 10;
long long fact[maxn];
vector<long long> get(long long n, set<long long> a) {
  vector<long long> ans;
  while (a.size() > 1) {
    long long m = a.size();
    long long ord = n / fact[m - 1];
    n -= ord * fact[m - 1];
    long long cnt = 0, x = -1;
    for (auto i : a) {
      if (cnt == ord) x = i;
      cnt++;
    }
    assert(x != -1);
    a.erase(x);
    ans.push_back(x);
  }
  ans.push_back(*a.begin());
  return ans;
}
long long a[maxn], ps[maxn];
int32_t main() {
  ios::sync_with_stdio(0), cin.tie(0);
  long long n, q;
  cin >> n >> q;
  for (long long i = 0; i < n; i++) a[i] = i + 1;
  for (long long i = 1; i <= n; i++) ps[i] = ps[i - 1] + a[i - 1];
  long long tot = 0;
  for (long long i = 0; i < maxn; i++) fact[i] = (!i ? 1 : fact[i - 1] * i);
  for (long long i = 0; i < q; i++) {
    long long tp;
    cin >> tp;
    if (tp == 2) {
      long long x;
      cin >> x;
      tot += x;
      long long l = max(0LL, n - M);
      set<long long> s;
      for (long long j = l; j < n; j++) s.insert(a[j]);
      vector<long long> v = get(tot, s);
      for (long long j = l; j < n; j++) a[j] = v[j - l];
      for (long long j = l + 1; j <= n; j++) ps[j] = ps[j - 1] + a[j - 1];
    } else {
      long long l, r;
      cin >> l >> r;
      cout << ps[r] - ps[l - 1] << '\n';
    }
  }
}