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 kth;
long long orig[200001];
long long psum[200001];
long long fact[21];
bool visited[21];
int remain[21];
int fixedNum[21];
int bound;
void initVisited() {
  for (int i = 0; i < 21; i++) visited[i] = 0;
}
void fixNum() {
  long long curK = kth;
  initVisited();
  for (int i = 1; i <= bound; i++) {
    int curPos = (curK / fact[(bound - i)]) + 1;
    int cnt = 0;
    for (int j = 1; j <= bound; j++) {
      if (!visited[j]) {
        cnt++;
        if (cnt == curPos) {
          visited[j] = true;
          fixedNum[i] = remain[j];
          break;
        }
      }
    }
    curK -= fact[(bound - i)] * ((long long)curPos - 1);
  }
}
long long getSum(long long b, long long e) { return psum[e] - psum[b - 1]; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  fact[0] = 1;
  for (int i = 1; i < 21; i++) fact[i] = fact[i - 1] * i;
  kth = 0;
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) orig[i] = i;
  for (int i = 1; i <= n; i++) psum[i] = psum[i - 1] + orig[i];
  bound = min(n, 20);
  for (int i = 1; i <= bound; i++) remain[i] = n - bound + i;
  for (int q = 0; q < k; q++) {
    int ty;
    cin >> ty;
    if (ty == 1) {
      int l, r;
      cin >> l >> r;
      fixNum();
      long long sum = getSum(l, r);
      int updateL = max(l, n - bound + 1);
      int updateR = min(r, n);
      long long diff = 0;
      for (int i = updateL; i <= updateR; i++) {
        diff += i - fixedNum[i - (n - bound)];
      }
      sum -= diff;
      cout << sum << '\n';
    } else if (ty == 2) {
      int step;
      cin >> step;
      kth += step;
    }
  }
}