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 JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class Task5 {
    long[] fact;

    public static void main(String[] args) throws IOException {
        new Task5().solve();
    }

    ArrayList<Integer> getLastOrder(ArrayDeque<Integer> lastElements, ArrayList<Integer> order, int pref, long skip) {
        if (pref < 0) return order;

        ArrayList<Integer> removed = new ArrayList<Integer>();
        int curr = lastElements.removeFirst();
        while (fact[pref] <= skip) {
            skip -= fact[pref];
            removed.add(curr);
            curr = lastElements.removeFirst();
        }

        for (int i = removed.size() - 1; i >= 0; i--) {
            lastElements.addFirst(removed.get(i));
        }

        order.add(curr);
        getLastOrder(lastElements, order, pref - 1, skip);
        return order;
    }

    private void solve() throws IOException {
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);

        StringTokenizer tokenizer = new StringTokenizer(f.readLine());
        int n = Integer.parseInt(tokenizer.nextToken());
        int q = Integer.parseInt(tokenizer.nextToken());

        fact = new long[20];
        fact[0] = 1;
        for (int i = 1; i < 20; i++) {
            fact[i] = fact[i - 1] * i;
        }

        ArrayDeque<Integer> lastElements = new ArrayDeque<Integer>();
        for (int i = n; n - i < Math.min(15, n); i--) lastElements.addFirst(i);
        long absChange = 0;
        ArrayList<Integer> order = getLastOrder(lastElements.clone(), new ArrayList<Integer>(), lastElements.size() - 1, 0);
        for (int q1 = 0; q1 < q; q1++) {
            tokenizer = new StringTokenizer(f.readLine());
            int command = Integer.parseInt(tokenizer.nextToken());
            if (command == 1) {
                int l = Integer.parseInt(tokenizer.nextToken());
                int r = Integer.parseInt(tokenizer.nextToken());
                long res = 0;
                if (l > n - lastElements.size()) {
                    l -= n - lastElements.size() + 1;
                    r -= n - lastElements.size() + 1;
                    for (int i = l; i <= r; i++) {
                        res += order.get(i);
                    }
                } else {
                    int mid = Math.min(r, n - lastElements.size());
                    mid = Math.max(l - 1, mid);
                    res += arithmeticProgression(l, mid);
                    int in = r - mid;
                    for (int i = 0; i < in; i++) res += order.get(i);
                }

                out.println(res);
            } else {
                int x = Integer.parseInt(tokenizer.nextToken());
                absChange += x;
                order = getLastOrder(lastElements.clone(), new ArrayList<Integer>(), lastElements.size() - 1, absChange);
            }
        }

        out.close();
    }

    static long arithmeticProgression(int from, int to) {
        if (to < from) return 0;
        return (long) (from + to) * (to - from + 1) / 2;
    }
}