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 long long maxN = 200009;
const long long inf = (long long)1e18 + 1;
long long N, Q;
long long a[maxN];
long long fact[20];
vector<long long> curr;
vector<long long> Kthperm(long long K) {
  long long n = min(N, (long long)15);
  vector<long long> v(n + 1, 1);
  long long i, j, k;
  long long fv = n - 1;
  vector<long long> ret;
  for (i = 1; i <= n; i++) {
    long long pos = 1;
    while (K > fact[fv]) {
      pos++;
      K -= fact[fv];
    }
    fv--;
    for (j = 1; j <= n; j++) {
      if (v[j] == 1) {
        pos--;
      }
      if (pos == 0 && v[j] == 1) {
        ret.push_back(j);
        v[j] = 0;
      }
    }
  }
  for (i = 0; i < n; i++) {
    if (N > 15) {
      ret[i] += (N - 15);
    }
  }
  return ret;
}
long long calc(long long l, long long r) {
  if (r < l) return 0;
  return ((r - l + 1) * (l + r)) / 2;
}
void solve() {
  long long i, j, k;
  cin >> N >> Q;
  fact[0] = 1;
  for (i = 1; i <= 16; i++) {
    fact[i] = fact[i - 1] * i;
  }
  curr = Kthperm(1);
  long long cp = 1;
  while (Q--) {
    long long type;
    cin >> type;
    if (type == 1) {
      long long l, r;
      cin >> l >> r;
      long long ans = calc(l, min(r, N - 15));
      for (i = max(N - 14, l); i <= r; i++) {
        long long p = i - 1;
        if (N > 15) {
          p -= (N - 15);
        };
        ans += curr[p];
      }
      cout << ans << "\n";
    } else {
      long long x;
      cin >> x;
      cp += x;
      curr = Kthperm(cp);
    }
  }
}
signed main() {
  if (true) {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  }
  long long T = 1;
  if (false) {
    cin >> T;
  }
  while (T--) {
    solve();
  }
}