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>
#pragma GCC optimize("O2")
using namespace std;
const double PI = atan(1.0) * 4;
const int di[4] = {-1, 0, 1, 0};
const int dj[4] = {0, -1, 0, 1};
const long long INF = (long long)2e18 + 50;
const int maximum = numeric_limits<int>::max();
const int minimum = numeric_limits<int>::min();
const int mod = 998244353;
long long fac[16];
void init() {
  fac[0] = 1;
  for (int i = 1; i <= 15; i++) {
    fac[i] = 1ll * fac[i - 1] * i;
  }
}
int binpow(int a, int b, int m) {
  a %= m;
  int res = 1;
  while (b > 0) {
    if (b & 1) res = 1ll * res * a % m;
    a = 1ll * a * a % m;
    b >>= 1;
  }
  return res;
}
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
long long gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
vector<long long> kth(const vector<long long> &R, long long k) {
  vector<long long> num = R, ret;
  for (int i = R.size() - 1; ~i; --i) {
    long long x = k / fac[i];
    ret.emplace_back(num[x]);
    num.erase(num.begin() + x);
    k %= fac[i];
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int T;
  T = 1;
  while (T--) {
    init();
    int n, q;
    cin >> n >> q;
    int num_right = min(n, 14);
    int num_left = n - num_right;
    vector<long long> psum(num_left + 1);
    vector<long long> R(num_right);
    iota(psum.begin() + 1, psum.end(), 1);
    iota(R.begin(), R.end(), num_left + 1);
    partial_sum(psum.begin(), psum.end(), psum.begin());
    long long k = 0;
    while (q--) {
      int x;
      cin >> x;
      if (x == 1) {
        int l, r;
        cin >> l >> r;
        long long ans = 0;
        if (l <= num_left) {
          ans += psum[min(num_left, r)] - psum[l - 1];
        }
        if (r > num_left) {
          auto vt = kth(R, k);
          for (int i = max(0, l - num_left - 1); i < r - num_left; ++i) {
            ans += vt[i];
          }
        }
        cout << ans << "\n";
      } else {
        int num;
        cin >> num;
        k += num;
      }
    }
  }
  return 0;
}