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 mod = 1e9 + 7;
long long fpow(long long a, long long b, long long p = mod) {
  long long ans = 1;
  a %= p;
  while (b) {
    if (b & 1) ans = ans * a % p;
    b >>= 1;
    a = a * a % p;
  }
  return ans;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int N = 1e5 + 7;
int n, q, v[20], p[20];
long long jc[20];
long long cal(int l, int r, int n, long long times) {
  for (int i = 1; i <= n; i++) v[i] = 1;
  for (int i = n, j = 1; i >= 1; i--, j++) {
    int cur = times / jc[i - 1], tmp = 0, pos;
    for (int k = 1; k <= n; k++) {
      tmp += v[k];
      if (tmp > cur) {
        pos = k;
        break;
      }
    }
    v[pos] = 0;
    p[j] = pos;
    times %= jc[i - 1];
  }
  long long ans = 0;
  for (int i = l; i <= r; i++) ans += p[i];
  return ans;
}
void solve() {
  jc[0] = 1;
  for (int i = 1; i <= 15; i++) jc[i] = jc[i - 1] * i;
  scanf("%d %d", &n, &q);
  long long res = 0;
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int l, r;
      scanf("%d %d", &l, &r);
      if (n < 15) {
        long long ans = cal(l, r, n, res);
        printf("%lld\n", ans);
      } else {
        long long ans = 0;
        if (r <= n - 15)
          ans += 1LL * (l + r) * (r - l + 1) / 2;
        else if (l <= n - 15)
          ans += 1LL * (n - 15 + l) * (n - 15 - l + 1) / 2 +
                 cal(1, r - n + 15, 15, res) + (r - n + 15) * (n - 15);
        else {
          ans += cal(l - n + 15, r - n + 15, 15, res) + (n - 15) * (r - l + 1);
        }
        printf("%lld\n", ans);
      }
    } else {
      int x;
      scanf("%d", &x);
      res += x;
    }
  }
}
int main() {
  int T = 1;
  while (T--) solve();
  return 0;
}