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 mod = 1e9 + 7;
const int maxn = 1e6 + 10;
const int maxm = 4e3 + 10;
const long long inf = 1e18;
int T, n, m;
long long a[maxn], len, f[20];
void update(long long x) {
  set<int> s;
  for (int i = n - len + 1; i <= n; i++) s.insert(i);
  long long r, num;
  for (int i = 1, j = n - len + 1; i <= len; i++, j++) {
    r = x / f[len - i];
    x %= f[len - i];
    num = 0;
    auto it = s.begin();
    while (1) {
      num++;
      if (num == r + 1) {
        a[j] = *it;
        s.erase(it);
        break;
      }
      it++;
    }
  }
}
long long query(int l, int r) {
  long long ans = 0;
  if (l <= n - len) ans = a[min(r * 1ll, n - len)] - a[l - 1];
  for (int i = max(n - len + 1, l * 1ll); i <= r; i++) ans += a[i];
  return ans;
}
int main() {
  scanf("%d%d", &n, &m);
  len = min(14, n);
  f[0] = 1;
  for (long long i = 1; i <= len; i++) f[i] = f[i - 1] * i;
  for (int i = 1; i <= n - len; i++) {
    a[i] = a[i - 1] + i;
  }
  for (int i = n - len + 1; i <= n; i++) a[i] = i;
  int id, l, r;
  long long now = 0;
  while (m--) {
    scanf("%d%d", &id, &l);
    if (id == 1) {
      scanf("%d", &r);
      printf("%lld\n", query(l, r));
    } else {
      now += l;
      update(now);
    }
  }
  return 0;
}