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;
void pr(long long x, bool fl = false) {
  cout << x << endl;
  if (fl) exit(0);
}
void pr(long long x, long long y, bool fl = false) {
  cout << x << " " << y << endl;
  if (fl) exit(0);
}
const long long mod = 998244353;
const long long inf = LLONG_MAX;
const long long minf = LLONG_MIN;
const long long maxn = 200000;
const long long maxc = 26;
void chkmax(long long &_max, long long x) { _max = max(_max, x); }
void chkmin(long long &_min, long long x) { _min = min(_min, x); }
void modadd(long long &now, long long x, long long p = mod) {
  while (now < 0) now += p;
  x %= p;
  now %= p;
  now += x;
  now %= p;
  while (now < 0) now += p;
}
void modmul(long long &now, long long x, long long p = mod) {
  while (now < 0) now += p;
  x %= p;
  now %= p;
  now *= x;
  now %= p;
  while (now < 0) now += p;
}
long long qpow(long long a, long long b, long long p = mod) {
  long long ans = 1;
  a %= p;
  while (b) {
    if (b & 1) ans *= a;
    a *= a;
    ans %= p;
    a %= p;
    b /= 2;
  }
  return ans;
}
long long inv(long long a, long long p = mod) { return qpow(a % p, p - 2, p); }
long long c, xx, y, z, n, m, q, l, r, u, v, w, tmp, nxt, rt, cur, dis;
long long ans, val, flag;
long long a[maxn + 5], f[maxn + 5], x;
long long peng(long long x) {
  if (x <= 1) return 1;
  return peng(x - 1) * x;
}
void change() {
  for (long long i = 1; i <= y; i++) f[i] = i, a[i] = 0;
  long long j = 1, now = x;
  while (j <= y) {
    long long k = now / peng(y - j);
    now = now % peng(y - j);
    long long cnt = 0, nowj = 1;
    while (nowj <= y) {
      while (a[nowj] == 1) nowj++;
      cnt++;
      if (cnt > k) break;
      nowj++;
    }
    f[j] = nowj;
    a[nowj] = 1;
    j++;
  }
}
signed main() {
  ios::sync_with_stdio(false);
  long long test_case = 1;
  long long opt = 0;
  while (test_case--) {
    cin >> n >> q;
    y = min(n, 15ll);
    for (long long i = 1; i <= y; i++) f[i] = i;
    if (n <= 15) {
      for (long long i = 1; i <= q; i++) {
        cin >> opt;
        if (opt == 1) {
          ans = 0;
          change();
          cin >> l >> r;
          for (long long i = l; i <= r; i++) ans += f[i];
          pr(ans);
        } else
          cin >> z, x += z;
      }
    } else {
      for (long long i = 1; i <= q; i++) {
        cin >> opt;
        if (opt == 1) {
          ans = 0;
          cin >> l >> r;
          if (r >= n - 14) {
            change();
            for (long long i = max(1ll, l - n + 15); i <= r - n + 15; i++)
              ans += n - 15 + f[i];
            r = n - 15;
          }
          if (l <= r) ans += (l + r) * (r - l + 1) / 2;
          pr(ans);
        } else {
          cin >> z;
          x += z;
        }
      }
    }
  }
}