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 N = 200005;
const long long B = 20;
long long JC[N];
long long n, q;
long long sum[N], a[N], b[N];
bool used[N];
void solve(long long m) {
  long long t = min(n, B);
  for (long long u = 1; u <= t + 5; u++) used[u] = false;
  for (long long u = 1; u <= t; u++) {
    long long x = t - u, t1 = m / JC[x];
    m = m - t1 * JC[x];
    for (long long i = 1; i <= t; i++) {
      if (used[i] == false) {
        if (t1 == 0) {
          b[u] = i;
          used[i] = true;
          break;
        } else
          t1--;
      }
    }
  }
  long long o = n - t + 1;
  for (long long u = o; u <= n; u++) a[u] = b[u - o + 1] + o - 1;
  for (long long u = o; u <= n; u++) sum[u] = sum[u - 1] + a[u];
}
int main() {
  JC[0] = 1;
  for (long long u = 1; u <= B; u++) JC[u] = JC[u - 1] * u;
  scanf("%lld%lld", &n, &q);
  for (long long u = 1; u <= n; u++) a[u] = u;
  for (long long i = 1; i <= n; i++) sum[i] = sum[i - 1] + i;
  long long m = 0;
  while (q--) {
    long long op;
    scanf("%lld", &op);
    if (op == 1) {
      long long l, r;
      scanf("%lld%lld", &l, &r);
      printf("%lld\n", sum[r] - sum[l - 1]);
    } else {
      long long x;
      scanf("%lld", &x);
      m = m + x;
      solve(m);
    }
  }
  return 0;
}