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;
long long fac[200005];
int position[200005];
long long pen[200005];
void update(int ind, int add) {
  while (ind < 200005) {
    pen[ind] += add;
    ind += ind & (-ind);
  }
}
long long query(int ind) {
  long long ret = 0;
  while (ind > 0) {
    ret = ret + pen[ind];
    ind = ind & (ind - 1);
  }
  return ret;
}
int cc[200005];
void updateC(int ind, int add) {
  while (ind < 200005) {
    cc[ind] += add;
    ind += ind & (-ind);
  }
}
void construct(int n, long long cur) {
  int l = min(100, n);
  for (int i = n - l + 1; i <= n; ++i) {
    update(position[i], -i);
    updateC(i, 1);
  }
  for (int i = n - l + 1; i <= n; ++i) {
    int co = n - i + 1;
    long long u = cur / fac[co - 1];
    assert(u < co);
    cur -= fac[co - 1] * u;
    ++u;
    long long sum = 0;
    int ind = 0;
    for (int j = 20; j >= 0; --j) {
      int nind = ind + (1 << j);
      if (nind <= n && sum + cc[nind] < u) {
        ind = nind;
        sum += cc[nind];
      }
    }
    ++ind;
    updateC(ind, -1);
    position[ind] = i;
    update(position[ind], ind);
  }
  assert(cur == 0);
}
void solve() {
  int n, q;
  scanf("%d %d ", &n, &q);
  fac[0] = 1;
  const long long inf = 1e13;
  for (int i = 1; i <= n; ++i) {
    fac[i] = min(inf, fac[i - 1] * i);
    position[i] = i;
    update(i, i);
  }
  long long cur = 0;
  while (q--) {
    int type;
    scanf("%d ", &type);
    if (type == 1) {
      int l, r;
      scanf("%d %d ", &l, &r);
      printf("%lld\n", query(r) - query(l - 1));
    } else {
      int x;
      scanf("%d  ", &x);
      cur += x;
      construct(n, cur);
    }
  }
}
int main() {
  solve();
  return 0;
}