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 Max = 2e5 + 5;
const int Mod = 998244353;
const int inf = 0x3f3f3f3f;
long long sum[Max], p[Max];
long long F[20];
void init() {
  F[0] = 1;
  for (int i = 1; i <= 15; i++) F[i] = F[i - 1] * i;
}
long long ConTor(int n, int Mid) {
  long long X = 1;
  for (int i = Mid + 1; i < n; i++) {
    long long temp = 0;
    for (int j = i + 1; j <= n; j++) {
      if (p[j] > p[i]) continue;
      temp++;
    }
    X += temp * F[n - i];
  }
  return X;
}
bool vis[Max];
void invConTor(int n, int Mid, long long X) {
  X--;
  long long rest = X;
  vector<int> vec;
  for (int i = Mid + 1; i <= n; i++) {
    vis[i] = 0;
    vec.push_back(i);
  }
  for (int i = Mid + 1; i <= n; i++) {
    int num = rest / F[n - i];
    rest = rest % F[n - i];
    p[i] = vec[num];
    vec.erase(vec.begin() + num);
  }
}
int main(void) {
  init();
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) p[i] = i;
  int Mid = max(0, n - 15);
  for (int i = 1; i <= Mid; i++) sum[i] = sum[i - 1] + p[i];
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int l, r;
      cin >> l >> r;
      if (r <= Mid)
        cout << sum[r] - sum[l - 1] << '\n';
      else if (l <= Mid) {
        long long temp = sum[Mid] - sum[l - 1];
        for (int i = Mid + 1; i <= r; i++) temp += p[i];
        cout << temp << '\n';
      } else {
        long long temp = 0;
        for (int i = l; i <= r; i++) temp += p[i];
        cout << temp << '\n';
      }
    } else {
      long long x;
      cin >> x;
      long long now = ConTor(n, Mid);
      now += x;
      invConTor(n, Mid, now);
    }
  }
}