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 maxn = 5e5 + 10;
const int mod = 1e9 + 7;
long long f[50], pre[maxn], sum;
void init() {
  f[0] = 1;
  for (int i = 1; i <= 15; i++) f[i] = f[i - 1] * i;
}
int n, q, m;
vector<int> v;
long long solve(int x) {
  long long ans = 0;
  int k = n - m;
  if (x <= k)
    ans = 1ll * x * (x + 1) / 2;
  else {
    long long cur = sum;
    x = x - k, v.clear();
    ans = 1ll * k * (k + 1) / 2 + 1ll * x * k;
    for (int i = 1; i <= m; i++) v.push_back(i);
    for (int i = m - 1; i >= m - x; i--) {
      int id = cur / f[i];
      ans += v[id];
      cur %= f[i];
      v.erase(v.begin() + id);
    }
  }
  return ans;
}
int main() {
  init(), m = 0;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) pre[i] = pre[i - 1] + i;
  while (q--) {
    int op, x, y;
    scanf("%d%d", &op, &x);
    if (op == 1) {
      scanf("%d", &y);
      long long ans = solve(y) - solve(x - 1);
      printf("%lld\n", ans);
    } else {
      sum += x;
      while (sum >= f[m]) m++;
    }
  }
}