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 N = 2e5 + 5;
const int M = 1e6 + 5;
const int INF = 0x3f3f3f3f;
const long long LL_MAX = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
inline long long read() {
  long long ans = 0;
  int f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = ans * 10 + c - '0';
    c = getchar();
  }
  return ans * f;
}
long long sum[N];
long long fac[20];
int n, q;
long long cnt;
void init() {
  for (int i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + i;
  }
  fac[0] = 1;
  for (int i = 1; i <= 17; i++) {
    fac[i] = fac[i - 1] * i;
  }
}
vector<int> Kthpermutation(int n, long long k) {
  k++;
  vector<int> vec;
  vector<int> res;
  for (int i = 0; i < n; i++) vec.push_back(i + 1);
  if (k == 1) return vec;
  long long base = fac[n - 1];
  k--;
  for (int i = 0; i < n - 1; i++) {
    long long divres = k / base;
    k = k % base;
    int num = vec[divres];
    res.push_back(num);
    vec.erase(vec.begin() + divres);
    base = fac[n - 2 - i];
  }
  res.push_back(vec[0]);
  return res;
}
int main() {
  n = read(), q = read();
  init();
  int R = max(1, n - 15);
  while (q--) {
    int op = read();
    if (op == 1) {
      int l = read(), r = read();
      if (r < R) {
        printf("%lld\n", sum[r] - sum[l - 1]);
      } else {
        long long ans = 0;
        int start;
        if (l < R) {
          ans += sum[R - 1] - sum[l - 1];
          start = R;
        } else {
          start = l;
        }
        vector<int> v;
        v = Kthpermutation(n - (R - 1), cnt);
        for (int i = start - R; i <= r - R; i++) {
          ans = ans + v[i] + (R - 1);
        }
        printf("%lld\n", ans);
      }
    } else {
      int x;
      scanf("%d", &x);
      cnt += x;
    }
  }
  return 0;
}