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 MAX_N = 2e5 + 7;
long long fac[MAX_N] = {1ll,         1ll,          2ll,          6ll,
                        24ll,        120ll,        720ll,        5040ll,
                        40320ll,     362880ll,     3628800ll,    39916800ll,
                        479001600ll, 6227020800ll, 87178291200ll};
vector<long long> a, now;
long long n, q, tot = 14;
void init() {
  for (long long i = 0; i < tot; ++i) a.push_back(i + 1), now.push_back(i + 1);
}
vector<long long> change(long long x, vector<long long> a, long long n) {
  if (n == 1) return a;
  long long pos = x / fac[n - 1], mod = x % fac[n - 1];
  vector<long long> nxt, res;
  for (long long i = 0; i < n; ++i) {
    if (i == pos) continue;
    nxt.push_back(a[i]);
  }
  nxt = change(mod, nxt, n - 1);
  res.push_back(a[pos]);
  for (long long i = 0; i < n - 1; ++i) res.push_back(nxt[i]);
  return res;
}
inline long long sum(long long x) { return (x + 1) * x / 2; }
long long get_sum(long long l, long long r) {
  long long res = 0, dec;
  if (n <= tot) {
    dec = tot - n;
    for (long long i = l + dec; i <= r + dec; ++i) {
      res += now[i - 1] - dec;
    }
  } else {
    if (l <= n - tot) {
      res -= sum(l - 1);
      if (r <= n - tot) {
        res += sum(r);
      } else {
        res += sum(n - tot);
        for (long long i = 1; i <= r - n + tot; ++i) {
          res += now[i - 1] + n - tot;
        }
      }
    } else {
      for (long long i = l - n + tot; i <= r - n + tot; ++i) {
        res += now[i - 1] + n - tot;
      }
    }
  }
  return res;
}
signed main() {
  init();
  scanf("%lld%lld", &n, &q);
  long long add = 0;
  while (q--) {
    long long opt, x, l, r;
    scanf("%lld", &opt);
    if (opt == 1) {
      scanf("%lld%lld", &l, &r);
      printf("%lld\n", get_sum(l, r));
    } else {
      scanf("%lld", &x);
      add += x;
      now = change(add, a, tot);
    }
  }
  return 0;
}