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 poww(long long a, long long b, long long md) {
  return (!b ? 1
             : (b & 1 ? a * poww(a * a % md, b / 2, md) % md
                      : poww(a * a % md, b / 2, md) % md));
}
const long long MAXN = 1e6 + 10;
const long long INF = 8e18;
const long long MOD = 1e9 + 7;
const long long FACT = 17;
long long fact[FACT], k = 1, q, n;
inline vector<int> perm_t(int n, long long k, long long L = 1) {
  vector<int> ans, dig;
  for (int i = 1; i <= n; i++) dig.push_back(L + i - 1);
  for (int i = 1; i <= n; i++) {
    long long red = fact[n - i];
    long long d = (k + red - 1) / red;
    k -= (d - 1) * red;
    auto it = dig.begin() + d - 1;
    ans.push_back(*it);
    dig.erase(it);
  }
  return ans;
}
inline long long sum(long long L, long long R) {
  return R * (R + 1) / 2 - L * (L - 1) / 2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  fact[0] = fact[1] = 1;
  for (int i = 2; i < FACT; i++) fact[i] = fact[i - 1] * i;
  cin >> n >> q;
  while (q--) {
    int cmd;
    cin >> cmd;
    if (cmd == 1) {
      long long l, r;
      cin >> l >> r;
      if (n < FACT) {
        vector<int> P = perm_t(n, k);
        int ans = 0;
        for (int i = l - 1; i < r; i++) ans += P[i];
        cout << ans << '\n';
      } else {
        vector<int> P = perm_t(FACT, k, n - FACT + 1);
        long long ans = 0;
        if (l < n - FACT + 1) ans += sum(l, min(n - FACT, r));
        for (int i = max(n - FACT + 1, l); i <= r; i++)
          ans += P[i - (n - FACT + 1)];
        cout << ans << '\n';
      }
    } else {
      int x;
      cin >> x;
      k += x;
    }
  }
  return 0;
}