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;
int n, m, q, op, x, y, c[16];
long long sum[200001], jc[16] = {1}, now;
inline int lowbit(int x) { return x & (-x); }
void update(int x, int v) {
  for (; x <= m; x += lowbit(x)) c[x] += v;
}
int Find(int x) {
  int now = 0;
  for (register int i = 1 << 4; i; i >>= 1)
    if (now + i <= m && c[now + i] < x) x -= c[now + i], now += i;
  return now + 1;
}
int main() {
  scanf("%d%d", &n, &q);
  m = min(n, 15);
  for (register int i = 1; i <= 15; ++i) jc[i] = jc[i - 1] * i;
  for (register int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + i;
  if (n <= 15) {
    while (q--) {
      scanf("%d", &op);
      if (op == 2) {
        scanf("%d", &x);
        now += x;
      } else {
        scanf("%d%d", &x, &y);
        long long ans = 0, tmp = now;
        for (register int i = 1; i <= m; ++i) update(i, 1);
        for (register int i = 1; i <= m; ++i) {
          int z = tmp / jc[m - i] + 1;
          tmp %= jc[m - i];
          int pos = Find(z);
          if (i >= x && i <= y) ans += pos;
          update(pos, -1);
        }
        printf("%lld\n", ans);
      }
    }
    return 0;
  }
  while (q--) {
    scanf("%d", &op);
    if (op == 2) {
      scanf("%d", &x);
      now += x;
    } else {
      scanf("%d%d", &x, &y);
      long long ans = 0;
      if (y <= n - 15) {
        printf("%lld\n", sum[y] - sum[x - 1]);
        continue;
      }
      if (x <= n - 15) {
        ans += sum[n - 15] - sum[x - 1];
        x = n - 14;
      }
      long long tmp = now;
      for (register int i = 1; i <= m; ++i) update(i, 1);
      for (register int i = 1; i <= m; ++i) {
        int z = tmp / jc[m - i] + 1;
        tmp %= jc[m - i];
        int pos = Find(z);
        if (i >= x - n + 15 && i <= y - n + 15) ans += pos;
        update(pos, -1);
      }
      printf("%lld\n", ans + (long long)(n - 15) * (y - x + 1));
    }
  }
  return 0;
}