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 int mxN = 2e5 + 5;
int n, q, a[mxN];
long long fct[mxN];
void get_perm(long long x) {
  for (int i = max(0, n - 14); i < n; i++) {
    a[i] = i + 1;
  }
  for (int i = max(0, n - 14); i < n - 1; i++) {
    while (x >= fct[n - i - 1]) {
      if (x == 0) return;
      int ch = n + 1;
      for (int j = i + 1; j < n; j++) {
        if (ch > a[j] && a[j] > a[i]) {
          ch = a[j];
        }
      }
      vector<int> vct;
      for (int j = i; j < n; j++) {
        if (a[j] != ch) vct.push_back(a[j]);
      }
      sort(vct.begin(), vct.end());
      a[i] = ch;
      int k = 0;
      for (int j = i + 1; j < n; j++) {
        a[j] = vct[k++];
      }
      x -= fct[n - i - 1];
    }
  }
}
void solve() {
  for (int i = 0; i < n; i++) {
    a[i] = i + 1;
  }
  fct[0] = 1;
  for (int i = 1; i <= n + 2; i++) {
    fct[i] = fct[i - 1] * i;
    if (fct[i] >= 1LL * mxN * mxN) break;
  }
  long long t, l, r;
  long long perm = 0;
  while (q--) {
    cin >> t;
    if (t == 1) {
      cin >> l >> r;
      l--;
      r--;
      get_perm(perm);
      long long right_border = max(0, n - 14);
      long long sm = 0;
      if (n < 14) {
        for (int i = l; i <= r; i++) {
          sm += a[i];
        }
      } else {
        for (int i = max(right_border, l); i <= r; i++) {
          sm += a[i];
        }
        r = min(r, right_border - 1);
        if (l <= n - 14) {
          sm += 1LL * (l + r + 2) * (r - l + 1) / 2;
        }
      }
      cout << sm << "\n";
    } else {
      cin >> l;
      perm += l;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  solve();
  return 0;
}