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;
const int INF = 1e9 + 7;
const int mod = 1e9 + 7;
long long int powmod(long long int a, long long int b) {
  long long int res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
int nxt() {
  int x;
  cin >> x;
  return x;
}
class FenwickTree {
 private:
  vector<long long int> ft;

 public:
  FenwickTree() {}
  FenwickTree(int n) { ft.assign(n + 1, 0LL); }
  long long int rsq(int b) {
    long long int sum = 0;
    for (; b; b -= (b & (-b))) sum += ft[b];
    return sum;
  }
  long long int rsq(int a, int b) { return rsq(b) - (a == 1 ? 0 : rsq(a - 1)); }
  void adjust(int k, long long int v) {
    for (; k < (int)ft.size(); k += (k & (-k))) ft[k] += v;
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  vector<long long int> fact;
  fact.push_back(1);
  for (long long int i = 2; fact.back() < 10e10; i++) {
    fact.push_back(fact.back() * i);
  }
  int n = nxt(), q = nxt();
  FenwickTree ft(n);
  for (int i = 1; i <= n; i++) ft.adjust(i, i);
  long long int sum = 0;
  vector<int> a(n);
  iota((a).begin(), (a).end(), 1);
  while (q--) {
    int k = nxt();
    if (k == 1) {
      int l = nxt(), r = nxt();
      printf("%lld\n", ft.rsq(l, r));
    } else {
      long long int x;
      cin >> x;
      sum += x;
      long long int oldsum = sum;
      int i = fact.size() - 1;
      vector<int> c;
      while (sum < fact[i]) i--;
      int len = i + 2;
      for (int j = 0; j < len; j++) {
        c.push_back(a.back());
        ft.adjust(n - j, -a.back());
        a.pop_back();
      }
      assert(c.size() < 30);
      sort((c).begin(), (c).end());
      while (i >= 1) {
        assert(sum >= 0);
        int t = sum / fact[i];
        assert(0 <= t && t < c.size());
        a.push_back(c[t]);
        c.erase(c.begin() + t);
        sum %= fact[i];
        i--;
      }
      assert(c.size() == 2);
      assert(sum == 0 || sum == 1);
      if (sum) {
        a.push_back(c[1]);
        a.push_back(c[0]);
      } else {
        a.push_back(c[0]);
        a.push_back(c[1]);
      }
      for (int i = 0; i < len; i++) {
        ft.adjust(n - i, a[n - i - 1]);
      }
      sum = oldsum;
    }
  }
  return 0;
}