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 MOD1 = 1e9 + 7;
const int MOD2 = 998244353;
const long long INF = 2 * 1e18;
const long double PI = 3.14159265358979323846;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long fpow(long long a, long long b, long long m) {
  if (!b) return 1;
  long long ans = fpow(a * a % m, b / 2, m);
  return (b % 2 ? ans * a % m : ans);
}
long long inv(long long a, long long m) { return fpow(a, m - 2, m); }
long long a[20], sum[200005], p[20], vis[25];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, q, tmp = 0;
  cin >> n >> q;
  p[0] = 1;
  for (long long i = 1; i <= n; i++) sum[i] = sum[i - 1] + i;
  for (long long i = 1; i <= 16; i++) p[i] = p[i - 1] * i;
  for (long long i = 1; i <= 16; i++) a[i] = max(0LL, n - 16) + i;
  while (q--) {
    long long x;
    cin >> x;
    if (x == 1) {
      long long l, r;
      cin >> l >> r;
      long long k = max(0LL, n - 16);
      if (k < 0) {
        cout << sum[r] - sum[l - 1] << "\n";
      } else {
        if (k < l) {
          long long ans = 0;
          for (int i = l - k; i <= r - k; i++) ans += a[i];
          cout << ans << "\n";
        } else if (r < k)
          cout << sum[r] - sum[l - 1] << "\n";
        else {
          long long ans = 0;
          for (int i = 1; i <= r - k; i++) ans += a[i];
          cout << sum[k] - sum[l - 1] + ans << "\n";
        }
      }
    } else {
      long long t, k = max(0LL, n - 16);
      cin >> t;
      tmp += t;
      long long cur = tmp;
      for (int i = min(16LL, n); i >= 1; i--) {
        long long cnt = 1;
        while (vis[cnt]) cnt++;
        while (cur >= p[i - 1]) {
          cnt++;
          while (vis[cnt]) cnt++;
          cur -= p[i - 1];
        }
        vis[cnt] = 1;
        a[min(16LL, n) - i + 1] = cnt + k;
      }
      for (long long i = 1; i <= 20; i++) vis[i] = 0;
    }
  }
}