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 maxn = 2e5 + 5;
const int maxA = 15;
int maxa;
int a[maxn], b[maxA], c[maxA];
long long fact[maxA + 1];
void precompute() {
  fact[0] = 1;
  for (int i = 1; i <= maxA; ++i) {
    fact[i] = fact[i - 1] * i;
  }
}
void getPermutation(long long k, int n) {
  for (int i = 0; i < maxa; ++i) {
    c[i] = i;
    for (int j = 1; j <= maxa; ++j) {
      if (j * fact[maxa - i - 1] > k) {
        b[i] = j - 1;
        k -= (j - 1) * fact[maxa - i - 1];
        break;
      }
    }
  }
  for (int i = 0; i < maxa; ++i) {
    int j = 0, cnt = 0;
    while (j < maxa && cnt < b[i] + 1) {
      cnt += (c[j] != -1);
      ++j;
    }
    b[i] = a[n - maxa + c[j - 1] + 1];
    c[j - 1] = -1;
  }
}
long long SUM(int n) { return n * 1ll * (n + 1) / 2; }
void solve() {
  precompute();
  int n, q;
  cin >> n >> q;
  maxa = min(maxA, n);
  for (int i = 1; i <= n; ++i) {
    a[i] = i;
  }
  for (int i = n - maxa + 1; i <= n; ++i) {
    b[i + maxa - n - 1] = a[i];
  }
  long long permute = 0;
  int flg = 0;
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int L, R;
      cin >> L >> R;
      long long sum = 0;
      if (R <= n - maxa) {
        sum = SUM(R) - SUM(L - 1);
      } else {
        sum = SUM(n - maxa) - SUM(min(n - maxa, L - 1));
        if (flg) {
          getPermutation(permute, n);
          flg = 0;
        }
        for (int i = max(L, n - maxa + 1); i <= R; ++i) {
          sum += b[i + maxa - n - 1];
        }
      }
      cout << sum << '\n';
    } else {
      int x;
      cin >> x;
      permute += x;
      flg = 1;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}