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 int p = 1e9 + 7;
long long int fact[20];
void fac() {
  fact[0] = 1;
  for (long long int i = 1; i <= 15; i++) {
    fact[i] = i * fact[i - 1];
  }
}
void solve(int n, long long int k, vector<long long int> &v,
           vector<long long int> &res) {
  if (fact[n] == k) {
    reverse(v.begin(), v.end());
    for (int i = 0; i < v.size(); i++) {
      res.push_back(v[i]);
    }
    return;
  }
  if (k % fact[n - 1] == 0) {
    long long int idx = k / fact[n - 1];
    res.push_back(v[idx - 1]);
    v.erase(v.begin() + idx - 1);
    solve(n - 1, fact[n - 1], v, res);
  } else {
    long long int idx = k / fact[n - 1] + 1;
    res.push_back(v[idx - 1]);
    v.erase(v.begin() + idx - 1);
    solve(n - 1, k % fact[n - 1], v, res);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  t = 1;
  fac();
  while (t--) {
    long long int n;
    long long int q;
    cin >> n >> q;
    vector<long long int> v1;
    long long int cnt = 0;
    long long int val = n;
    while (cnt < 15 && cnt < n) {
      cnt++;
      v1.push_back(val);
      val--;
    }
    reverse(v1.begin(), v1.end());
    long long int count = 1;
    while (q--) {
      if (n <= 15) {
        int x;
        cin >> x;
        if (x == 1) {
          int l, r;
          cin >> l >> r;
          vector<long long int> v;
          for (int i = 0; i < n; i++) {
            v.push_back(i + 1);
          }
          vector<long long int> res;
          solve(n, count, v, res);
          l--;
          r--;
          long long int ans = 0;
          for (int i = l; i <= r; i++) {
            ans += res[i];
          }
          cout << ans << '\n';
          ;
        }
        if (x == 2) {
          int l;
          cin >> l;
          count += l;
        }
      } else {
        long long int x;
        cin >> x;
        if (x == 1) {
          long long int l, r;
          cin >> l >> r;
          if (r <= (n - 15)) {
            long long int ans1 = (r * (r + 1)) / 2;
            long long int ans2 = (l * (l + 1)) / 2;
            cout << (ans1 - ans2 + l) << '\n';
            ;
          } else if (l > (n - 15)) {
            vector<long long int> v;
            v = v1;
            vector<long long int> res;
            long long int s = n - 14;
            l -= s;
            r -= s;
            solve(15, count, v, res);
            long long int ans = 0;
            for (int i = l; i <= r; i++) {
              ans += res[i];
            }
            cout << ans << '\n';
            ;
          } else {
            long long int ans = ((n - 15) * (n - 14)) / 2;
            vector<long long int> v;
            v = v1;
            vector<long long int> res;
            solve(15, count, v, res);
            int i = n - 14;
            int j = 0;
            while (i <= r) {
              ans += res[j];
              j++;
              i++;
            }
            l--;
            ans -= ((l) * (l + 1)) / 2;
            cout << ans << '\n';
            ;
          }
        }
        if (x == 2) {
          int l;
          cin >> l;
          count += l;
        }
      }
    }
  }
  return 0;
}