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 f[16];
long long sum = 0;
long long n, q;
long long a[16];
void incantor(long long k, long long n) {
  long long x;
  vector<long long> v;
  for (long long i = 1; i <= n; i++) v.push_back(i);
  for (long long i = 1; i < n; ++i) {
    a[i] = v[(x = k / f[n - i])];
    v.erase(v.begin() + x);
    k %= f[n - i];
  }
  a[n] = v[0];
}
void print() {}
int main() {
  f[0] = 1;
  for (long long i = 1; i <= 15; i++) f[i] = f[i - 1] * i;
  scanf("%lld%lld", &n, &q);
  if (n <= 15) {
    while (q--) {
      long long opt;
      scanf("%lld", &opt);
      if (opt == 1) {
        long long l, r;
        scanf("%lld%lld", &l, &r);
        long long ans = 0;
        incantor(sum, n);
        for (long long i = l; i <= r; i++) ans += a[i];
        printf("%lld\n", ans);
      } else {
        long long x;
        scanf("%lld", &x);
        sum += x;
      }
    }
  } else {
    while (q--) {
      long long opt;
      scanf("%lld", &opt);
      if (opt == 1) {
        long long l, r;
        scanf("%lld%lld", &l, &r);
        incantor(sum, 15);
        long long ans = 0;
        if (r >= n - 14) {
          long long st = n - 14;
          if (l >= n - 14) {
            for (long long i = l; i <= r; i++)
              ans += a[i - (n - 15)] + (n - 15);
          } else {
            for (long long i = n - 15 + 1; i <= r; i++)
              ans += a[i - (n - 15)] + (n - 15);
            ans += (n - 15 - l + 1) * (n - 15 + l) / 2;
          }
        } else {
          ans += (l + r) * (r - l + 1) / 2;
        }
        printf("%lld\n", ans);
      } else {
        long long x;
        scanf("%lld", &x);
        sum += x;
      }
    }
  }
  return 0;
}