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 mod = 998244353;
int t, n, q;
long long x;
long long sum[1000010];
int a[1000010];
long long f[1000010];
set<int> se;
void solve(int pos, long long re) {
  if (re == 0) {
    for (auto it = se.begin(); it != se.end(); ++it) {
      a[pos++] = *it;
    }
    return;
  }
  for (int i = 1; i <= 15; ++i) {
    if (f[i] - 1 >= re) {
      for (int j = pos; j <= n - i; ++j) {
        a[j] = *se.begin();
        se.erase(se.begin());
      }
      auto it = se.begin();
      for (int j = 1; j <= re / f[i - 1]; ++j) {
        it++;
      }
      int val = *it;
      a[n - i + 1] = val;
      it = se.lower_bound(val);
      se.erase(it);
      solve(n - i + 2, re - (re / f[i - 1]) * f[i - 1]);
      return;
    }
  }
}
int main() {
  f[0] = 1;
  for (int i = 1; i <= 15; ++i) {
    f[i] = f[i - 1] * i;
  }
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) {
    a[i] = i;
    sum[i] = sum[i - 1] + i;
  }
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int l, r;
      scanf("%d%d", &l, &r);
      for (int i = max(1, n - 16); i <= n; ++i) {
        sum[i] = sum[i - 1] + a[i];
      }
      printf("%lld\n", sum[r] - sum[l - 1]);
    } else {
      int xx;
      scanf("%d", &xx);
      for (int i = max(1, n - 16); i <= n; ++i) {
        a[i] = i;
      }
      x += xx;
      se.clear();
      for (int i = 1; i <= 15; ++i) {
        if (f[i] - 1 >= x) {
          for (int j = n; j >= n - i + 1; --j) {
            se.insert(j);
          }
          a[n - i + 1] = n - i + 1 + x / f[i - 1];
          auto it = se.lower_bound(a[n - i + 1]);
          se.erase(it);
          solve(n - i + 2, x - (x / f[i - 1]) * f[i - 1]);
          break;
        }
      }
    }
  }
  return 0;
}