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 = 5e5 + 1, M = 17;
const long long INF = 1e18 + 5;
inline long long read() {
  long long sum = 0, fh = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fh = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    sum = sum * 10 + c - '0';
    c = getchar();
  }
  return sum * fh;
}
inline int read2() {
  char c = getchar();
  while (c < '0' || c > '9') {
    c = getchar();
  }
  return c - '0';
}
inline void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
inline int gcd(int x, int y) { return y == 0 ? x : gcd(y, x % y); }
inline long long ab(long long x) { return x < 0 ? -x : x; }
long long jc[M];
int sy[M] = {0}, pp[M];
inline void pl(long long x, int y) {
  for (int i = 1; i <= y; ++i) sy[i] = 0;
  for (int i = y - 1; i >= 1; --i) {
    long long z = x / jc[i], js = 0;
    for (int j = 1; j <= y; ++j) {
      if (!sy[j]) ++js;
      if (js == z + 1) {
        js = j;
        break;
      }
    }
    x %= jc[i];
    pp[y - i] = js;
    sy[js] = 1;
  }
  for (int j = 1; j <= y; ++j) {
    if (!sy[j]) {
      pp[y] = j;
      break;
    }
  }
}
long long a[N];
const long long p = 998244353;
int main() {
  jc[0] = 1;
  for (long long i = 1; i <= 15; ++i) jc[i] = jc[i - 1] * i;
  int n = read(), m = read();
  long long sx = 0;
  for (int i = 1; i <= n; ++i) a[i] = i;
  for (int i = 1; i <= n; ++i) a[i] += a[i - 1];
  if (n > 14) {
    int n2 = n - 14;
    for (int i = 1; i <= m; ++i) {
      int o = read(), l = read(), r;
      if (o == 2) {
        sx += l;
        continue;
      }
      r = read();
      long long ans = 0;
      if (r > n2) {
        if (l <= n2) ans = a[n2] - a[l - 1], l = n2 + 1;
        pl(sx, 14);
        l -= n2, r -= n2;
        for (int j = l; j <= r; ++j) ans += pp[j] + n2;
      } else
        ans = a[r] - a[l - 1];
      write(ans), putchar('\n');
    }
  } else {
    for (int i = 1; i <= m; ++i) {
      int o = read(), l = read(), r;
      if (o == 2) {
        sx += l;
        continue;
      }
      r = read();
      long long ans = 0;
      pl(sx, n);
      for (int j = l; j <= r; ++j) ans += pp[j];
      write(ans), putchar('\n');
    }
  }
}