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;
long long read() {
  long long kkk = 0, x = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') c = getchar(), x = -1;
  while (c >= '0' && c <= '9')
    kkk = (kkk << 3) + (kkk << 1) + (c - '0'), c = getchar();
  return kkk * x;
}
long long n, q, a[200001], p[200001], sum[10], tree[200001], Tree[800001];
void xg(long long zone, long long v) {
  while (zone <= n) {
    tree[zone] += v;
    zone += (zone & -zone);
  }
}
long long cx(long long zone) {
  long long bck = 0;
  while (zone) {
    bck += tree[zone];
    zone -= (zone & -zone);
  }
  return bck;
}
void XG(long long l, long long r, long long zone, long long v, long long bj) {
  if (l == r) {
    Tree[bj] += v;
    return;
  }
  long long mid = (l + r) / 2;
  if (zone <= mid)
    XG(l, mid, zone, v, bj * 2);
  else
    XG(mid + 1, r, zone, v, bj * 2 + 1);
  Tree[bj] = Tree[bj * 2] + Tree[bj * 2 + 1];
}
long long CX(long long l, long long r, long long K, long long bj) {
  if (l == r) return l;
  long long mid = (l + r) / 2;
  if (Tree[bj * 2] >= K)
    return CX(l, mid, K, bj * 2);
  else
    return CX(mid + 1, r, K - Tree[bj * 2], bj * 2 + 1);
}
signed main() {
  n = read(), q = read();
  sum[0] = 1;
  for (long long i = 1; i <= 9; ++i) sum[i] = sum[i - 1] * i;
  for (long long i = 1; i <= n; ++i) a[i] = i, p[i] = 0, xg(i, i);
  while (q--) {
    long long type = read();
    if (type == 1) {
      long long l = read(), r = read();
      printf("%lld\n", cx(r) - cx(l - 1));
    } else {
      long long x = read(), len = min(n - 1, 9ll), rest = 0;
      for (long long i = len; i >= 0; --i) {
        p[n - i] += x / sum[i];
        x %= sum[i];
      }
      for (long long i = 0; i <= len; ++i) {
        p[n - i] += rest;
        rest = p[n - i] / (i + 1);
        p[n - i] %= (i + 1);
        if (i == len && rest) ++len;
      }
      for (long long i = 0; i <= len; ++i)
        xg(n - i, -a[n - i]), XG(1, n, a[n - i], 1, 1);
      for (long long i = len; i >= 0; --i) {
        a[n - i] = CX(1, n, p[n - i] + 1, 1);
        xg(n - i, a[n - i]);
        XG(1, n, a[n - i], -1, 1);
      }
    }
  }
  return 0;
}