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;
const long long mod = 998244353;
int n, q;
long long a[411100];
long long b[411100];
long long sum[411100];
long long pr[411100];
int get_num(set<int>& ft, int idx) {
  idx--;
  auto it = ft.begin();
  while (idx-- > 0) {
    it++;
  }
  int ret = *it;
  ft.erase(it);
  return ret;
}
vector<int> proc(long long tot) {
  vector<int> ans;
  set<int> ft;
  int len = 0;
  for (int i = 1; i <= 15; ++i) {
    ft.insert(i);
    if (pr[i] >= tot && len == 0) {
      len = i;
    }
  }
  for (int i = len; i > 0; --i) {
    int idx = (tot + pr[i - 1] - 1) / pr[i - 1];
    int x = get_num(ft, idx);
    if (x > 0) {
      ans.push_back(x);
    }
    tot %= pr[i - 1];
    if (tot == 0) tot = pr[i - 1];
  }
  return ans;
}
long long sol(long long tot, int l, int r) {
  tot++;
  vector<int> ret = proc(tot);
  int len = ret.size();
  for (int i = n + 1 - len, j = 0; i <= n; ++i, ++j) {
    ret[j] += n - len;
    b[i] = b[i - 1] + ret[j];
  }
  long long ans = b[r] - b[l - 1];
  for (int i = n + 1 - len; i <= n; ++i) {
    b[i] = sum[i];
  }
  return ans;
}
int main() {
  cin >> n >> q;
  memset(sum, 0, sizeof(sum));
  pr[0] = 1;
  for (int i = 1; i <= 15; ++i) {
    pr[i] = pr[i - 1] * i;
  }
  for (int i = 1; i <= n; ++i) {
    a[i] = i;
    sum[i] = sum[i - 1] + a[i];
    b[i] = sum[i];
  }
  long long tot = 0;
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int l, r;
      cin >> l >> r;
      cout << sol(tot, l, r) << endl;
    } else {
      int t;
      cin >> t;
      tot += t;
    }
  }
}