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>
const int N = 2e5 + 10;
const int mod = 998244353;
const double PI = acos(-1.0);
const double eps = 1e-8;
using namespace std;
inline int read() {
  int num = 0, w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) {
    num = (num << 3) + (num << 1) + (ch ^ 48);
    ch = getchar();
  }
  return w ? -num : num;
}
inline int lowbit(int x) { return x & -x; }
int n, q, t, a[N];
long long pre[N], sumx, cnt[16];
void mynext_permutation() {
  long long temp = sumx;
  int m = upper_bound(cnt, cnt + 16, sumx) - cnt;
  int b[m + 1], c[m + 1], len = m;
  for (int i = 1; i <= m; ++i) c[i] = i;
  long long suf[16];
  suf[m] = 1;
  for (int j = 2; j <= m; ++j) suf[m + 1 - j] = suf[m + 2 - j] * j;
  for (int i = 1; i < m && temp; ++i) {
    int chos = temp / suf[i + 1] + 1;
    temp = temp % suf[i + 1];
    b[i] = c[chos];
    for (int i = chos; i < len; ++i) c[i] = c[i + 1];
    --len;
  }
  for (int i = len, j = m; i >= 1; --i, --j) b[j] = c[i];
  for (int i = n - m, j = 1; i < n; ++i, ++j) a[i] = n - m + b[j];
}
void solve() {
  cin >> n >> q;
  for (int i = 0; i < n; ++i) a[i] = i + 1;
  pre[0] = a[0];
  for (int i = 1; i < n; ++i) pre[i] = pre[i - 1] + a[i];
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int l, r;
      cin >> l >> r;
      --l, --r;
      cout << pre[r] - pre[l] + a[l] << endl;
    } else {
      int x;
      cin >> x;
      sumx += x;
      mynext_permutation();
      pre[0] = a[0];
      for (int i = max(1, n - 15); i < n; ++i) pre[i] = pre[i - 1] + a[i];
    }
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cnt[1] = 1;
  for (int i = 2; i <= 15; ++i) cnt[i] = cnt[i - 1] * i;
  solve();
  return 0;
}