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>
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(0);
  int n, q;
  std::cin >> n >> q;
  const int L = std::min(n, 15);
  std::vector<long long> fac(L);
  fac[0] = 1;
  for (int i = 1; i < L; ++i) {
    fac[i] = fac[i - 1] * i;
  }
  auto calc = [&](long long x, int l, int r) {
    std::vector<int> a(L);
    for (int i = L - 1; i >= 0; --i) {
      a[i] = x / fac[i];
      x %= fac[i];
    }
    std::reverse(a.begin(), a.end());
    std::vector<bool> vis(L);
    std::vector<int> p(L);
    for (int i = 0; i < L; ++i) {
      int t = 0;
      p[i] = 0;
      while (vis[p[i]] || t < a[i]) {
        t += !vis[p[i]];
        ++p[i];
      }
      vis[p[i]] = true;
    }
    int res = 0;
    for (int i = l; i < r; ++i) {
      res += p[i];
    }
    return res;
  };
  long long sum = 0;
  while (q--) {
    int op;
    std::cin >> op;
    if (op == 1) {
      int l, r;
      std::cin >> l >> r;
      --l;
      if (r <= n - L) {
        std::cout << 1ll * (l + r + 1) * (r - l) / 2 << "\n";
      } else if (l <= n - L) {
        std::cout << 1ll * (l + n - L + 1) * (n - L - l) / 2 +
                         1ll * (n - L + 1) * (r - n + L) +
                         calc(sum, 0, r - n + L)
                  << "\n";
      } else {
        std::cout << 1ll * (n - L + 1) * (r - l) +
                         calc(sum, l - n + L, r - n + L)
                  << "\n";
      }
    } else {
      int x;
      std::cin >> x;
      sum += x;
    }
  }
}