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;
int n, q;
long long fact[20];
int v[20];
long long sp[200005];
void getpermutation(long long sz, long long k) {
  vector<bool> used(sz + 15, 0);
  for (int i = 1; i <= sz; i++) v[i] = i;
  for (int i = 1; i <= sz; i++) {
    for (int j = 1; j <= sz; j++) {
      if (used[j] == 0) {
        if (k > fact[sz - i]) {
          k -= fact[sz - i];
        } else {
          used[j] = 1;
          v[i] = j;
          break;
        }
      }
    }
  }
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  fact[0] = 1;
  for (int i = 1; i <= 15; i++) {
    fact[i] = 1LL * fact[i - 1] * 1LL * i;
  }
  for (int i = 1; i <= n; i++) sp[i] = sp[i - 1] + 1LL * i;
  if (n <= 15) {
    for (int i = 1; i <= n; i++) v[i] = i;
    long long nrperm = 1;
    for (int i = 1; i <= q; i++) {
      int type;
      cin >> type;
      if (type == 1) {
        int l, r;
        cin >> l >> r;
        int sum = 0;
        for (int i = l; i <= r; i++) {
          sum += v[i];
        }
        cout << sum << '\n';
      } else {
        int x;
        cin >> x;
        nrperm += x;
        getpermutation(n, nrperm);
      }
    }
  } else {
    long long nrperm = 1;
    for (int i = 1; i <= 15; i++) {
      v[i] = i;
    }
    for (int i = 1; i <= q; i++) {
      int type;
      cin >> type;
      if (type == 2) {
        long long x;
        cin >> x;
        nrperm += x;
        getpermutation(15, nrperm);
        for (int j = n - 14; j <= n; j++) {
          sp[j] = sp[j - 1] + 1LL * v[j - (n - 15)] + (n - 15);
        }
      } else {
        int l, r;
        cin >> l >> r;
        cout << sp[r] - sp[l - 1] << '\n';
      }
    }
  }
}