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;
template <class T>
void read(T& x) {
  T res = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    res = (res << 3) + (res << 1) + c - '0';
    c = getchar();
  }
  x = res * f;
}
const long long N = 200000 + 10;
const int mod = 1e9 + 7;
long long n, q, sum[N], tot = 1, fac[22], len;
vector<int> find(long long x) {
  x--;
  vector<int> res, ans;
  for (int i = 1; i <= len; i++) res.push_back(i);
  for (int i = 1; i <= len; i++) {
    long long re = x / fac[len - i];
    ans.push_back(res[re]);
    res.erase(res.begin() + (re));
    x %= fac[len - i];
  }
  return ans;
}
int main() {
  read(n);
  read(q);
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + i;
  fac[0] = 1;
  for (int i = 1; i <= 20; i++) fac[i] = fac[i - 1] * i;
  long long L = max(1ll, n - 15);
  len = n - L + 1;
  while (q--) {
    int ty;
    read(ty);
    if (ty == 1) {
      int l, r;
      read(l), read(r);
      if (r < L) {
        printf("%lld\n", sum[r] - sum[l - 1]);
      } else {
        long long ans = 0;
        int st;
        if (l < L) {
          ans += sum[L - 1] - sum[l - 1];
          st = L;
        } else {
          st = l;
        }
        vector<int> pt = find(tot);
        for (int i = st; i <= r; i++) ans += L - 1 + pt[i - L];
        printf("%lld\n", ans);
      }
    } else {
      int x;
      read(x);
      tot += x;
    }
  }
  return 0;
}