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 maxn = 2e5 + 100;
const long long fac = 6227020800;
long long last[20];
long long sum[maxn];
long long x;
int bot;
void init(int n) {
  sum[0] = 0;
  for (int i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + i;
  }
  for (int i = 1; i <= 14; i++) {
    last[i] = i;
  }
  x = 0;
  bot = n - 14;
}
int f[20];
void solve1(int l, int r) {
  long long res = 0;
  if (bot < 0) {
    for (int i = l; i <= r; i++) {
      res += last[i - bot] + bot;
    }
  } else {
    res += sum[min(r, bot)] - sum[min(l - 1, bot)];
    for (int i = max(1, l - bot); i <= r - bot; i++) {
      res += last[i] + bot;
    }
  }
  cout << res << endl;
}
void solve2(long long a) {
  x += a;
  memset(f, 0, sizeof f);
  long long now = x, it = fac, loc = 0, x = 0;
  while (loc < 13) {
    loc++;
    x = now / it + 1;
    now = now % it;
    it /= (14 - loc);
    for (int i = 1; i <= 14; i++) {
      x -= 1 ^ f[i];
      if (x == 0) {
        last[loc] = i;
        f[i] = 1;
        break;
      }
    }
  }
  for (int i = 1; i <= 14; i++) {
    if (f[i] == 0) {
      last[loc + 1] = i;
      f[i] = 1;
      break;
    }
  }
}
int main() {
  int q, n;
  int a, l, r;
  long long dx;
  cin >> n >> q;
  init(n);
  while (q--) {
    cin >> a;
    if (a == 1)
      cin >> l >> r, solve1(l, r);
    else
      cin >> dx, solve2(dx);
  }
  return 0;
}