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 maxn = 1e6 + 7, inf = 0x3f3f3f3f, mod = 1e9 + 7;
long long fac[20], a[maxn], t[maxn];
int lowbit(int x) { return x & (-x); }
void add(int x, int v) {
  while (x < maxn) {
    t[x] += v;
    x += lowbit(x);
  }
}
long long ask(int x) {
  long long res = 0;
  while (x) {
    res += t[x];
    x -= lowbit(x);
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  fac[0] = 1;
  for (int i = 1; i <= 15; ++i) fac[i] = fac[i - 1] * i;
  int n, q;
  cin >> n >> q;
  vector<int> st;
  int flag = inf;
  for (int i = 1; i <= n; ++i) {
    a[i] = i;
    add(i, i);
    if (n - i + 1 <= 15) {
      flag = min(i, flag);
      st.push_back(i);
    }
  }
  long long now = 0;
  int op, l, r, x;
  int c = 0;
  while (q--) {
    cin >> op;
    if (op == 1) {
      cin >> l >> r;
      cout << ask(r) - ask(l - 1) << '\n';
      c++;
    } else {
      cin >> x;
      now += x;
      vector<int> tmp = st;
      int sz = tmp.size();
      vector<bool> vis(sz, 0);
      long long cur = now;
      for (int i = flag; i <= n; ++i) {
        int kth = cur / fac[n - i], cnt = 0;
        if (i == n) assert(cur == 0);
        cur %= fac[n - i];
        for (int j = 0; j < sz; ++j) {
          if (vis[j]) continue;
          if (cnt == kth) {
            add(i, -a[i]);
            a[i] = tmp[j];
            add(i, a[i]);
            vis[j] = 1;
            break;
          }
          cnt++;
        }
      }
    }
  }
  return 0;
}