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;
using namespace std;
inline int read() {
  char ch;
  bool sign = 0;
  int res = 0;
  while (!isdigit(ch = getchar()))
    if (ch == '-') sign = 1;
  while (isdigit(ch)) res = res * 10 + (ch ^ 48), ch = getchar();
  if (sign)
    return -res;
  else
    return res;
}
const int maxn = 200000 + 10;
const long long inf = 2e18;
const int maxm = 200;
const long long mod = 998244353;
int maxlen = 14;
long long fac[maxm];
int n, q;
inline long long query(long long Rank, int l, int r) {
  if (r <= n - maxlen) {
    return 1LL * (l + r) * (r - l + 1) / 2;
  }
  long long ans = 0;
  vector<int> p = vector<int>(maxlen, -1);
  vector<int> vis = vector<int>(maxlen, 0);
  for (int i = 0; i < maxlen; ++i) {
    for (int j = 0; j < maxlen; ++j) {
      if (vis[j]) continue;
      if (Rank > fac[maxlen - i - 1]) {
        Rank -= fac[maxlen - i - 1];
      } else {
        p[i] = j;
        vis[j] = 1;
        break;
      }
    }
  }
  for (auto i = (0); i < (maxlen); ++i)
    assert(p[i] >= 0), p[i] += n - maxlen + 1;
  if (l <= n - maxlen) ans += 1LL * (l + n - maxlen) * (n - maxlen - l + 1) / 2;
  for (auto i = (max(l - 1, n - maxlen)); i < (r); ++i)
    ans += p[i - (n - maxlen)];
  return ans;
}
inline void solve() {
  long long Rank = 1;
  for (auto o = (1); o <= (q); ++o) {
    int t, l, r;
    scanf("%d", &(t));
    scanf("%d", &(l));
    if (t == 1)
      scanf("%d", &(r)), printf("%lld\n", query(Rank, l, r));
    else
      Rank += l;
  }
}
int main(void) {
  fac[0]++;
  for (auto i = (1); i <= (15); ++i) fac[i] = fac[i - 1] * i;
  scanf("%d%d", &n, &q);
  maxlen = min(n, maxlen);
  solve();
  return 0;
}