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 = 2e5 + 10;
int num[20];
long long sum[maxn], fac[20];
int bit[20];
int n, q;
inline int lowbit(int x) { return x & -x; }
void add(int pos, int val) {
  while (pos <= 15) {
    bit[pos] += val;
    pos += lowbit(pos);
  }
}
int query(int pos) {
  int ans = 0;
  while (pos) {
    ans += bit[pos];
    pos -= lowbit(pos);
  }
  return ans;
}
void solve(long long x) {
  iota(num, num + 20, 0);
  --x;
  memset(bit, 0, sizeof(bit));
  for (int i = 1; i <= 15; ++i) add(i, 1);
  for (int i = 14; i >= 0; --i) {
    int ls = i ? x / fac[i] : 0;
    int l = 1, r = 15, mid, cur;
    while (l <= r) {
      mid = l + r >> 1;
      if (query(mid) - 1 >= ls)
        cur = mid, r = mid - 1;
      else
        l = mid + 1;
    }
    num[15 - i] = cur;
    add(cur, -1);
    x %= fac[i];
  }
  if (n <= 15) {
    int delta = 15 - n;
    for (int i = 1, j = 16 - n; i <= n; ++i, ++j)
      sum[i] = sum[i - 1] + num[j] - delta;
  } else {
    int delta = n - 15;
    for (int i = n - 14, j = 1; i <= n; ++i, ++j)
      sum[i] = sum[i - 1] + num[j] + delta;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> q;
  fac[0] = 1;
  for (int i = 1; i <= 15; ++i) fac[i] = fac[i - 1] * i;
  for (int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + i;
  long long add = 1;
  while (q--) {
    int c, l, r, x;
    cin >> c;
    if (c == 1) {
      cin >> l >> r;
      cout << sum[r] - sum[l - 1] << '\n';
    } else {
      cin >> x;
      add += x;
      solve(add);
    }
  }
  return 0;
}