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 power(long long a, long long b) {
  long long res = 1;
  a = a % 1000000007;
  while (b) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b /= 2;
  }
  return res;
}
long long invmod(long long a) { return power(a, 1000000007 - 2); }
const int mxn = 2e5 + 10;
long long sum[mxn], per[mxn], n;
void setsum(int start) {
  for (int i = start; i <= n; i++) {
    sum[i] = sum[i - 1] + per[i];
  }
}
void kth(long long k) {
  long long cnt = 1, p = 1;
  for (; cnt * p <= k; p++) {
    cnt *= p;
  }
  int start, idx;
  start = idx = n - p + 1;
  vector<int> all;
  for (int i = n - p + 1; i <= n; i++) {
    all.push_back(i);
  }
  cout.flush();
  while (idx <= n) {
    int j = k / cnt;
    k %= cnt;
    per[idx++] = all[j];
    all.erase(all.begin() + j);
    if (p > 1) {
      cnt /= (--p);
    }
  }
  setsum(start);
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    int q;
    long long k = 0;
    cin >> n >> q;
    for (int i = 1; i <= n; i++) {
      per[i] = i;
      sum[i] = per[i] + sum[i - 1];
    }
    while (q--) {
      int x;
      cin >> x;
      if (x == 1) {
        int l, r;
        cin >> l >> r;
        cout << sum[r] - sum[l - 1] << "\n";
        cout.flush();
      } else {
        cin >> x;
        k += x;
        kth(k);
      }
    }
  }
  return 0;
}