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, pre[1000000], a[1000000], f[100], vis[100], ans, rec[100],
    temp[100];
int main() {
  cin >> n >> q;
  for (long long i = n; i >= 1; i--) pre[i] = pre[i + 1] + n - i + 1;
  f[0] = 1;
  for (long long i = 1; i <= 20; i++) f[i] = f[i - 1] * i;
  long long num = 0;
  for (long long o = 1; o <= q; o++) {
    long long tag;
    cin >> tag;
    if (tag == 1) {
      long long l, r;
      cin >> l >> r;
      if (n - r + 1 > 20) {
        cout << pre[n - r + 1] - pre[n - l + 2] << endl;
        continue;
      }
      long long t = num;
      for (long long i = min(n, (long long)20); i >= 1; i--)
        if (t >= f[i - 1]) {
          rec[i] = t / f[i - 1] + 1;
          t = t % f[i - 1];
        } else
          rec[i] = 1;
      memset(vis, 0, sizeof vis);
      long long base = n - 20;
      if (n <= 20) base = 0;
      for (long long i = min(n, (long long)20); i >= 1; i--) {
        long long xx, cnt = 0, now = 1;
        while (cnt < rec[i]) {
          if (!vis[now]) cnt++;
          now++;
        }
        now--;
        temp[i] = now + base;
        vis[now] = 1;
      }
      for (int i = min(n, (long long)20); i >= 1; i--)
        pre[i] = pre[i + 1] + temp[i];
      cout << pre[n - r + 1] - pre[n - l + 2] << endl;
    } else {
      long long x;
      cin >> x;
      num += x;
    }
  }
}