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 long long N = 2e5 + 5;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long sum, n, q;
long long a[16];
long long P[16];
long long fac[16];
bool vis[16];
void get(long long sum) {
  long long res = 0;
  for (long long i = 1; i < n; i++) res = (res + a[i]) * (n - i);
  for (long long i = 1; i <= 15; i++) {
    a[i] = sum / fac[15 - i];
    sum %= fac[15 - i];
  }
  memset(vis, 0, sizeof(vis));
  for (long long i = 1; i <= 15; i++) {
    long long cnt = 0;
    for (long long j = 1; j <= 15; j++) {
      if (!vis[j]) cnt++;
      if (cnt == a[i] + 1) {
        P[i] = j;
        vis[j] = 1;
        break;
      }
    }
  }
}
signed main() {
  n = read();
  q = read();
  fac[0] = 1;
  for (long long i = 1; i <= 15; i++) fac[i] = fac[i - 1] * i;
  while (q--) {
    long long opt;
    opt = read();
    if (opt == 1) {
      long long l, r;
      l = read();
      r = read();
      long long ans = 0;
      get(sum);
      if (r > n - 15) {
        if (l > n - 15) {
          for (long long i = l - n + 15; i <= r - n + 15; i++) ans += P[i];
          ans += (r - l + 1) * (n - 15);
        } else {
          ans += (l + n - 15) * (n - 14 - l) / 2;
          for (long long i = 1; i <= r - n + 15; i++) ans += P[i];
          ans += (r - n + 15) * (n - 15);
        }
      } else
        ans = (l + r) * (r - l + 1) / 2;
      printf("%lld\n", ans);
    }
    if (opt == 2) {
      long long x;
      x = read();
      sum += x;
    }
  }
  return 0;
}