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;
long long ord = 1;
long long fac[20];
int nums[20];
bool v[20];
int a[20];
int len;
long long presum;
long long getsum(int m) {
  if (m <= 0) return 0;
  if (m <= n - len) {
    return (1 + m) * (long long)m / 2LL;
  }
  long long ret = presum;
  for (int i = 1; i <= m - (n - len); ++i) ret += a[i];
  return ret;
}
long long updiv(long long x, long long y) {
  if (x == 0) return 1;
  return (x - 1) / y + 1;
}
void work() {
  long long o = ord;
  for (int i = 1; i <= len; ++i) v[i] = 0;
  for (int i = 1; i < len; ++i) {
    long long res = updiv(o, fac[len - i]);
    long long rp = res;
    int j;
    for (j = 1; j <= len; ++j) {
      if (!v[j]) {
        --res;
        if (!res) break;
      }
    }
    v[j] = 1;
    a[i] = nums[j];
    o -= fac[len - i] * (rp - 1);
  }
  for (int j = 1; j <= len; ++j)
    if (!v[j]) a[len] = nums[j];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  fac[1] = 1;
  for (int i = 2; i <= 15; ++i) fac[i] = fac[i - 1] * i;
  cin >> n >> q;
  if (n <= 15) {
    len = n;
    for (int i = 1; i <= n; ++i) a[i] = nums[i] = i;
    presum = 0;
  } else {
    len = 15;
    for (int i = 1; i <= 15; ++i) a[i] = nums[i] = n - 15 + i;
    presum = (n - 14) * (long long)(n - 15) / 2LL;
  }
  for (int i = 1; i <= q; ++i) {
    int x;
    cin >> x;
    if (x == 1) {
      int l, r;
      cin >> l >> r;
      cout << getsum(r) - getsum(l - 1) << endl;
    } else {
      cin >> x;
      ord += x;
      work();
    }
  }
}