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;
int n, q;
bool vis[15];
long long a[200100], s[200100];
int c;
int x, L, R;
long long t, dt;
long long f[15];
long long ans;
int find_order(int order) {
  for (int i = 0; i < 15; i++) {
    if (vis[i] == 0) continue;
    if (order == 0) return i;
    order--;
  }
}
void upd_a(int pos, int val) {
  a[pos] = val;
  if (pos) {
    s[pos] = s[pos - 1] + val;
  } else
    s[pos] = val;
}
int main() {
  f[0] = 1;
  for (int i = 1; i < 15; i++) f[i] = f[i - 1] * (long long)i;
  while (~scanf("%d%d", &n, &q)) {
    for (int i = 0; i < n; i++) {
      a[i] = i;
    }
    t = 0;
    s[0] = a[0];
    for (int i = 1; i < n; i++) s[i] = s[i - 1] + a[i];
    while (q--) {
      if (n < 15) {
        cin >> c;
        if (c == 1) {
          cin >> L >> R;
          L--;
          R--;
          if (L == 0)
            ans = s[R];
          else
            ans = s[R] - s[L - 1];
          ans += (R - L + 1);
          cout << ans << endl;
        } else {
          cin >> dt;
          memset(vis, 1, sizeof(vis));
          t += dt;
          int idx, order;
          long long tmp = t;
          for (int i = 0; i < n; i++) {
            order = tmp / f[n - 1 - i];
            tmp %= f[n - 1 - i];
            idx = find_order(order);
            vis[idx] = false;
            upd_a(i, idx);
          }
        }
      } else {
        cin >> c;
        if (c == 1) {
          cin >> L >> R;
          L--;
          R--;
          if (L == 0)
            ans = s[R];
          else
            ans = s[R] - s[L - 1];
          ans += (R - L + 1);
          cout << ans << endl;
        } else {
          cin >> dt;
          memset(vis, 1, sizeof(vis));
          t += dt;
          int idx, order;
          long long tmp = t;
          for (int i = 0; i < 15; i++) {
            order = tmp / f[14 - i];
            tmp %= f[14 - i];
            idx = find_order(order);
            vis[idx] = false;
            upd_a(n - 15 + i, idx + n - 15);
          }
        }
      }
    }
  }
  return 0;
}