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>
#pragma GCC optimize(2)
using namespace std;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3fLL;
const double PI = acos(-1.0);
const long double eps = 1e-8;
const int mod = 998244353;
const int N = 2e5 + 100;
const int M = 500 + 10;
const long long mm = (1LL << 32);
template <class T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  bool f = 0;
  for (; !isdigit(c); c = getchar()) f ^= c == '-';
  for (; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
  x = f ? -x : x;
}
template <class T>
inline void write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  T y = 1;
  int len = 1;
  for (; y <= x / 10; y *= 10) ++len;
  for (; len; --len, x %= y, y /= 10) putchar(x / y + 48);
}
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    b >>= 1;
    a = (a * a) % mod;
  }
  return ans;
}
long long sum[N], fac[N];
int vis[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  fac[0] = 1;
  for (int i = 1; i <= 17; i++) fac[i] = fac[i - 1] * i;
  int n, q;
  cin >> n >> q;
  int pos = max(1, n - 15);
  long long cur = 1;
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + i;
  while (q--) {
    int op, l, r;
    cin >> op;
    if (op == 1) {
      cin >> l >> r;
      printf("%lld\n", sum[r] - sum[l - 1]);
    } else {
      cin >> l;
      cur += l;
      for (int i = pos; i <= n; i++) vis[i] = 0;
      long long tmp = cur;
      for (int i = pos; i <= n; i++) {
        long long ret = 0;
        for (int j = pos; j <= n; j++) {
          if (!vis[j]) {
            ret += fac[n - i];
            if (ret >= tmp) {
              tmp -= ret - fac[n - i];
              sum[i] = sum[i - 1] + j;
              vis[j] = 1;
              break;
            }
          }
        }
      }
    }
  }
  return 0;
}