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 = 998244353;
const double PI = acos(-1.0);
inline long long gcd(long long a, long long b) {
  while (b != 0) {
    long long t = b;
    b = a % b;
    a = t;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long inv(long long a, long long m) {
  if (a == 1) return 1;
  return inv(m % a, m) * (m - m / a) % m;
}
inline long long ksm(long long a, long long x) {
  long long ret = 1;
  while (x) {
    if (x & 1) ret = (ret * a) % mod;
    a = (a * a) % mod;
    x >>= 1;
  }
  return ret;
}
const int MAXN = 5e5 + 10;
const int INF = 0x3f3f3f3f;
int n, m;
long long f[20], a[20];
vector<int> vec;
inline void incantor(int m, long long k) {
  int x;
  vec.clear();
  if (n > 15) {
    for (int i = n - 14; i <= n; i++) vec.push_back(i);
  } else {
    for (int i = 1; i <= m; ++i) vec.push_back(i);
  }
  for (int i = 1; i < m; ++i) {
    a[i] = vec[(x = k / f[m - i])];
    vec.erase(vec.begin() + x);
    k %= f[m - i];
  }
  a[m] = vec[0];
}
void solve() {
  int x, l, r, ty;
  long long now = 0;
  while (m--) {
    scanf("%d", &ty);
    if (ty == 1) {
      scanf("%d%d", &l, &r);
      long long sum = 0;
      if (n <= 15) {
        now %= f[n];
        incantor(n, now);
        for (int i = l; i <= r; i++) sum += a[i];
      } else if (r <= n - 15) {
        sum = 1LL * (l + r) * (r - l + 1) / 2;
      } else if (l <= n - 15) {
        sum = 1LL * (l + n - 15) * (n - 15 - l + 1) / 2;
        l = n - 15 + 1;
        int id = n - 15;
        incantor(15, now);
        for (int i = l; i <= r; i++) sum += a[i - id];
      } else {
        int id = n - 15;
        incantor(15, now);
        for (int i = l; i <= r; i++) sum += a[i - id];
      }
      printf("%lld\n", sum);
    } else {
      scanf("%d", &x);
      now += x;
    }
  }
}
void init() {
  f[1] = 1;
  for (int i = 2; i <= 15; ++i) f[i] = f[i - 1] * i;
  scanf("%d%d", &n, &m);
}
int main() {
  int T = 1;
  while (T--) {
    init();
    solve();
  }
  return 0;
}