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 <typename T>
string to_string(const vector<T>& vc, int w) {
  if (vc.empty()) return "";
  if (w + 1 == vc.size()) return to_string(vc[w]);
  return to_string(vc[w]) + "," + to_string(vc, w + 1);
}
template <typename T>
string to_string(const vector<T>& vc) {
  return "{" + to_string(vc, 0) + "}";
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
class DebugStream {
} LOG;
template <typename T>
DebugStream& operator<<(DebugStream& s, const T&) {
  return s;
}
long long f[40];
void init() {
  f[0] = 1;
  for (long long i = 1; i < 16; i++) f[i] = (i * f[i - 1]);
}
vector<int> NumberToPermutation(long long x, int len) {
  vector<int> ans(len);
  vector<bool> taken(len, false);
  for (int i = 0; i < int(len); i++) {
    long long acc = 0;
    vector<int> options;
    for (int j = 0; j < int(len); j++)
      if (!taken[j]) options.push_back(j);
    int total = (int)options.size();
    for (int j = 0; j < int(total); j++) {
      long long nxt = acc + f[total - 1];
      if (x >= acc && x < nxt) {
        x -= acc;
        ans[i] = options[j];
        taken[ans[i]] = true;
        break;
      }
      acc = nxt;
    }
  }
  for (int i = 0; i < int(len); i++) ans[i]++;
  return ans;
}
long long sum(int i, long long x, int len) {
  vector<int> ans = NumberToPermutation(x, len);
  long long res = 0;
  for (int j = 0; j < int(i); j++) {
    res += ans[j];
  }
  return res;
}
long long xum(int i, long long x, int len, int n) {
  if (i <= n - len) {
    return 1ll * i * (i + 1) / 2;
  }
  long long ans =
      1ll * (n - len) * (n - len + 1) / 2 + 1ll * (n - len) * (i - (n - len));
  ans += sum(i - (n - len), x, len);
  return ans;
}
int main() {
  init();
  int n, q;
  scanf("%d %d", &n, &q);
  long long cur = 0;
  for (int _ = 0; _ < int(q); _++) {
    int t;
    scanf("%d", &t);
    if (t == 2) {
      long long x;
      scanf("%lld", &x);
      cur += x;
    } else {
      int l, r;
      scanf("%d %d", &l, &r);
      if (n <= 15) {
        printf("%lld\n", sum(r, cur, n) - sum(l - 1, cur, n));
      } else {
        int len = 15;
        printf("%lld\n", xum(r, cur, len, n) - xum(l - 1, cur, len, n));
      }
    }
  }
}