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 N = 1e6 + 5, M = 15, P = 1e9 + 7, INF = 0X3F3F3F3F;
const unsigned long long D = 743, mo = 966335533, mo1 = 966331133;
int n, q;
long long now, ans, sum[N], fac[22];
std::vector<int> a;
void cantor(long long x, int y) {
  std::vector<int> v;
  a.clear();
  for (int i = 1; i <= y; i++) v.push_back(n - y + i);
  for (int i = y; i >= 1; i--) {
    long long r = x % fac[i - 1];
    long long t = x / fac[i - 1];
    x = r;
    sort(v.begin(), v.end());
    a.push_back(v[t]);
    v.erase(v.begin() + t);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  fac[0] = 1;
  for (int i = 1; i <= 15; i++) fac[i] = 1LL * fac[i - 1] * i;
  cin >> n >> q;
  if (n <= M) {
    cantor(0, n);
    for (int i = 1, op, l, r, x; i <= q; i++) {
      cin >> op;
      if (op == 1) {
        cin >> l >> r;
        ans = 0;
        for (int j = l - 1; j < r; j++) ans += a[j];
        cout << ans << '\n';
      } else {
        cin >> x;
        now += x;
        cantor(now, n);
      }
    }
  } else {
    cantor(0, M);
    for (int i = 1; i <= n - M; i++) {
      sum[i] = sum[i - 1] + i;
    }
    for (int i = 1, op, l, r, x; i <= q; i++) {
      cin >> op;
      if (op == 1) {
        cin >> l >> r;
        ans = 0;
        if (l <= n - M) ans += sum[min(n - M, r)] - sum[l - 1];
        for (int j = max(n - M + 1, l); j <= r; j++) ans += a[j - n + M - 1];
        cout << ans << '\n';
      } else {
        cin >> x;
        now += x;
        cantor(now, M);
      }
    }
  }
  return 0;
}