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;
inline long long read() {
  long long f = 1, ans = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = ans * 10 + c - '0';
    c = getchar();
  }
  return f * ans;
}
const long long MAXN = 2e5 + 11;
long long N, fac[16], Q, sum;
vector<long long> calc(long long res) {
  vector<long long> Ans;
  Ans.clear();
  set<long long> s;
  set<long long>::iterator it;
  for (long long i = 1; i <= min(N, 15ll); i++) s.insert(i);
  for (long long i = 1; i <= min(N, 15ll); i++) {
    it = s.begin();
    long long x = *it;
    while (res >= fac[min(N, 15ll) - i]) {
      it++;
      res -= fac[min(N, 15ll) - i];
    }
    Ans.push_back(*it);
    s.erase(*it);
  }
  return Ans;
}
signed main() {
  N = read(), Q = read();
  fac[0] = 1;
  for (long long i = 1; i <= 15; i++) fac[i] = fac[i - 1] * i;
  while (Q--) {
    long long opt = read();
    if (opt == 1) {
      long long l = read(), r = read();
      if (N <= 15) {
        vector<long long> Ans = calc(sum);
        long long ss = 0, res = 0;
        for (auto v : Ans) {
          ++ss;
          if (ss >= l && ss <= r) res += v;
        }
        printf("%lld\n", res);
      } else {
        vector<long long> Ans = calc(sum);
        long long ps = N - 15 + 1, res = 0;
        for (auto v : Ans) {
          if (l <= ps && ps <= r) res += v + (N - 15);
          ps++;
        }
        r = min(r, N - 15);
        if (l <= r) res += ((l + r) * (r - l + 1)) / 2;
        printf("%lld\n", res);
      }
    } else
      sum += read();
  }
}