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 int M = 2e5 + 5;
const long long INF = 1e18;
const int MOD = 1e9 + 7;
long long fac[15];
long long pf[20];
vector<int> fin;
long long n, q;
void des(long long x) {
  int i = 14;
  while (i > 0) {
    pf[i] = x / fac[i];
    x %= fac[i--];
  }
  vector<int> pos;
  long long a = max(1LL, n - 14);
  for (int i = a; i < n + 1; i++) pos.push_back(i);
  fin.clear();
  for (int i = pos.size() - 1; i >= 0; i--) {
    fin.push_back(pos[pf[i]]);
    pos.erase(pos.begin() + pf[i]);
  }
}
long long sum(int x) {
  long long r = 0;
  for (int i = 0; i < x; i++) r += fin[i];
  return r;
}
void go() {
  cin >> n >> q;
  fac[0] = 1;
  for (long long i = 1; i < 15; i++) fac[i] = fac[i - 1] * i;
  des(0);
  long long cont = 0;
  while (q--) {
    int c;
    cin >> c;
    if (c == 2) {
      long long x;
      cin >> x;
      cont += x;
      des(cont);
      continue;
    }
    long long l, r;
    cin >> l >> r;
    if (n <= 15) {
      long long a = sum(l - 1);
      long long b = sum(r);
      cout << b - a << "\n";
    } else {
      long long m = n - 15;
      if (l > m) {
        long long a = sum(l - 1 - m);
        long long b = sum(r - m);
        cout << b - a << "\n";
      } else if (r < m + 1) {
        cout << ((r - l + 1) * (r + l)) / 2 << "\n";
      } else {
        long long a = sum(r - m);
        long long b = ((m + l) * (m - l + 1)) / 2;
        cout << a + b << "\n";
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  t = 1;
  while (t--) go();
  return 0;
}