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;
long long n, q;
inline vector<long long> fun(long long n1, long long k, vector<long long> v2) {
  vector<long long> res;
  long long base = 1;
  for (long long i = n1 - 1; i > 1; i--) {
    base *= i;
  }
  for (long long i = 0; i < n1 - 1; i++) {
    long long div = k / base;
    k = k % base;
    vector<long long>::iterator it;
    it = v2.begin() + div;
    res.push_back(*it);
    v2.erase(it);
    base /= n1 - i - 1;
  }
  res.push_back(v2[0]);
  return res;
}
int main() {
  scanf("%lld%lld", &n, &q);
  vector<long long> v, v1, v3;
  for (long long i = 0; i < n; ++i) {
    v1.push_back(i + 1);
  }
  if (n > 20) {
    for (long long i = n - 13; i <= n; ++i) {
      v.push_back(i);
      v3.push_back(i);
    }
  } else {
    for (long long i = 0; i < n; ++i) {
      v.push_back(i + 1);
      v3.push_back(i + 1);
    }
  }
  long long base = 1;
  if (n < 15) {
    for (long long i = 1; i <= n; ++i) {
      base *= i;
    }
  }
  long long sum = 0;
  int tot = 0;
  while (q--) {
    long long t;
    scanf("%lld", &t);
    if (t == 1) {
      long long l, r;
      scanf("%lld%lld", &l, &r);
      if (n <= 20) {
        long long ans = 0;
        for (long long i = l - 1; i < r; ++i) {
          ans += v[i];
        }
        printf("%lld\n", ans);
        tot++;
      } else {
        if (r >= n - 13) {
          long long ans;
          if (l > n - 14)
            ans = 0;
          else {
            long long a = min(l, n - 14), b = min(n - 14, r);
            ans = (1 + b) * b / 2 - (a - 1) * a / 2;
          }
          for (long long i = max(l, n - 13); i <= r; ++i) {
            ans += v[i - (n - 13)];
          }
          printf("%lld\n", ans);
          tot++;
        } else {
          long long ans = (l + r) * (r - l + 1) / 2;
          printf("%lld\n", ans);
          tot++;
        }
      }
    } else {
      long long k;
      scanf("%lld", &k);
      sum += k;
      if (n < 15) sum %= base;
      v = fun(min(n, 14LL), sum, v3);
    }
  }
  return 0;
}