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 MOD = 1e9 + 7;
const int MAXN = 3e5 + 10;
int dirx[4] = {0, 1, 0, -1}, diry[4] = {1, 0, -1, 0};
int n, q;
struct Node {
  int op, l, r;
} op[MAXN];
long long c[MAXN], fac[18], tmp[20];
int lowbit(int x) { return x & -x; }
void upd(int x, int v) {
  while (x <= n) {
    c[x] += v;
    x += lowbit(x);
  }
}
long long qry(int x) {
  long long ans = 0;
  while (x > 0) {
    ans += c[x];
    x -= lowbit(x);
  }
  return ans;
}
int main() {
  fac[1] = fac[0] = 1;
  for (int i = 2; i <= 17; ++i) fac[i] = fac[i - 1] * i;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= q; ++i) {
    scanf("%d", &op[i].op);
    if (op[i].op == 1)
      scanf("%d%d", &op[i].l, &op[i].r);
    else
      scanf("%d", &op[i].l);
  }
  if (n <= 15) {
    long long now = 1;
    for (int j = 1; j <= q; ++j) {
      if (op[j].op == 1) {
        long long tmpval = now;
        for (int i = 1; i < n; ++i) {
          if (tmpval <= fac[n - i])
            tmp[i] = 1;
          else {
            tmp[i] = (tmpval + (fac[n - i] - 1)) / fac[n - i];
            tmpval -= fac[n - i] * (tmp[i] - 1);
          }
        }
        tmp[n] = 1;
        vector<int> lim;
        for (int i = 1; i <= n; ++i) lim.push_back(i);
        for (int i = 1; i <= n; ++i) {
          int val = lim[tmp[i] - 1];
          lim.erase(lim.begin() + (tmp[i] - 1));
          tmp[i] = val;
          upd(i, tmp[i]);
        }
        long long ans = qry(op[j].r) - qry(op[j].l - 1);
        printf("%lld\n", ans);
        for (int i = 1; i <= n; ++i) upd(i, -tmp[i]);
      } else
        now += op[j].l;
    }
  } else {
    long long now = 1;
    for (int i = 1; i <= n - 15; ++i) upd(i, i);
    for (int j = 1; j <= q; ++j) {
      if (op[j].op == 1) {
        long long tmpval = now;
        for (int i = 1; i < 15; ++i) {
          if (tmpval <= fac[15 - i])
            tmp[i] = 1;
          else {
            tmp[i] = (tmpval + (fac[15 - i] - 1)) / fac[15 - i];
            tmpval -= fac[15 - i] * (tmp[i] - 1);
          }
        }
        tmp[15] = 1;
        vector<int> lim;
        for (int i = 1; i <= 15; ++i) lim.push_back(i);
        for (int i = 1; i <= 15; ++i) {
          int val = lim[tmp[i] - 1];
          lim.erase(lim.begin() + (tmp[i] - 1));
          tmp[i] = val;
          tmp[i] += n - 15;
          upd(i + n - 15, tmp[i]);
        }
        long long ans = qry(op[j].r) - qry(op[j].l - 1);
        printf("%lld\n", ans);
        for (int i = 1; i <= 15; ++i) upd(i + n - 15, -tmp[i]);
      } else
        now += op[j].l;
    }
  }
  return 0;
}