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;
template <class T>
void print(vector<T>& v) {
  for (T x : v) cout << x << " ";
  cout << "\n";
}
void inp(vector<long long int>& v, int n) {
  for (int i = 0; i < n; ++i) scanf("%lld ", &v[i]);
}
long long int fact[16];
int vis[17];
void build() {
  fact[0] = 1;
  for (int i = 1; i < 16; ++i) fact[i] = fact[i - 1] * i;
  return;
}
void permut(vector<int>& v, int n, long long int k) {
  int j = n - 1, a = 0;
  int x, xx;
  memset(vis, 0, 4 * (n + 1));
  while (k > 1) {
    x = (k + fact[j] - 1) / fact[j];
    xx = x;
    for (int i = 1; i <= n; ++i) {
      if (vis[i] == 0) {
        xx--;
        if (xx == 0) {
          vis[i] = 1;
          v[a++] = i;
          break;
        }
      }
    }
    k -= (x - 1) * (fact[j]);
    j--;
  }
  for (int i = 1; i <= n; ++i) {
    if (vis[i] == 0) v[a++] = i;
  }
  return;
}
void sol() {
  int n, q;
  scanf("%d %d", &n, &q);
  long long int k = 1;
  vector<int> v(15, 0);
  int m = min(15, n);
  int nm = n - m;
  vector<long long int> pre(n + 1, 0);
  pre[0] = 0;
  for (int i = 1; i <= nm; ++i) pre[i] = pre[i - 1] + i;
  int x, l, r;
  long long int res = 0;
  for (int i = 0; i < q; ++i) {
    scanf("%d", &x);
    if (x == 1) {
      scanf("%d %d", &l, &r);
      l--;
      permut(v, m, k);
      res = 0;
      for (int j = 0; j < m; ++j) {
        pre[nm + j + 1] = pre[nm + j] + nm + v[j];
      }
      res = pre[r] - pre[l];
      printf("%lld\n", res);
    } else {
      scanf("%d", &l);
      k += l;
    }
  }
  return;
}
int main() {
  build();
  sol();
  return 0;
}