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 long long dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
const long long dxx[8] = {-1, -1, 0, 1, 1, 1, 0, -1},
                dyy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
const long long mod = 1000000007;
const long long base = 311;
const long long N = 2e5 + 5;
long long n, q, a[N];
long long fact[20], f[N];
void gogo() {
  cin >> n >> q;
  long long cur = 0;
  fact[0] = 1;
  for (long long i = 1; i <= 15; ++i) fact[i] = fact[i - 1] * i;
  for (long long i = 1; i <= n; ++i) {
    f[i] = f[i - 1] + i;
  }
  while (q--) {
    long long t;
    cin >> t;
    if (t == 1) {
      long long l, r;
      cin >> l >> r;
      cout << f[r] - f[l - 1] << '\n';
    } else {
      long long x;
      cin >> x;
      cur += x;
      long long beg = max(1ll, n - 14);
      for (long long i = beg; i <= n; ++i) f[i] = 0;
      vector<long long> p;
      for (long long i = beg; i <= n; ++i) p.push_back(i);
      vector<long long> np;
      long long tmp = cur;
      while (tmp) {
        for (long long i = ((long long)(p).size()) - 1; i >= 0; --i) {
          if (i * fact[((long long)(p).size()) - 1] <= tmp) {
            tmp -= i * fact[((long long)(p).size()) - 1];
            np.push_back(p[i]);
            p.erase(p.begin() + i);
            break;
          }
          if (i == 0) {
            np.push_back(p[0]);
            p.erase(p.begin());
          }
        }
      }
      for (long long i : p) np.push_back(i);
      for (long long i = beg; i <= n; ++i) {
        a[i] = np[i - beg];
        f[i] = f[i - 1] + a[i];
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (fopen("sol"
            ".inp",
            "r")) {
    freopen(
        "sol"
        ".inp",
        "r", stdin);
    freopen(
        "sol"
        ".out",
        "w", stdout);
  }
  gogo();
}