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.util.*;
import java.io.*;
public class EdF {
	static long[] mods = {1000000007, 998244353, 1000000009};
	static long mod = mods[0];
	public static MyScanner sc;
    public static PrintWriter out;
    static long[] fact;
	public static void main(String[] omkar) throws Exception{
		// TODO Auto-generated method stub
 		sc = new MyScanner();
 		PrintWriter out = new PrintWriter(System.out);
 		int n = sc.nextInt();
 		int q = sc.nextInt();
 		int len = Math.min(n,  15);
 		int[] last15 = new int[15];
 		int index = 0;
 		for(int j = Math.max(1,  n-14);j<=n;j++){
 			last15[index] = j;
 			index++;
 		}
 		int[] o = new int[15];
 		int[] e = new int[15];
 		for(int j =0;j<15;j++){
 			o[j] = last15[j];
 			e[j] = last15[j];
 		}
 		fact = new long[16];
 		fact[0] = 1;
		for(int j= 1;j<16;j++){
			fact[j] = (long)(j)*fact[j-1];
		}
 		long totalmoves = 1;
 		for(int j = 0;j<q;j++){
 			int type = sc.nextInt();
 			if (type == 1){
 				int l = sc.nextInt();
 				int r = sc.nextInt();
 				long sum = 0;
 				if (l <= n-len && r <= n-len){
 					sum+= (long)(l+r)*(long)(r-l+1)/2L;
 				}
 				else if (l <= n-len){
 					sum+=(long)(l+n-len)*(long)(n-len-l+1)/2L;
 				}
 				for(int k = Math.max(0,  l-(n-len)-1);k<r-(n-len);k++)
 					sum+=(long)last15[k];
 				out.println(sum);
 			}
 			else{
 				int moves =sc.nextInt();
 				totalmoves+=(long)(moves);
 				kthPerm(o, totalmoves, 0, len);
 				for(int k=0;k<15;k++){
 					last15[k] = o[k];
 					o[k] = e[k];
 				}

 			}
 			
 		}
 		out.println();
	 		
 		out.close();	
 	}
	public static void kthPerm(int[] a, long k, int l, int n){
		if (k != 1){
			long index = l+(k-1)/fact[n-l-1];
			int temp = a[(int)(index)];
			for(int j = (int)index;j>l;j--)
				a[j] = a[j-1];
			a[l] = temp;
			kthPerm(a, (k-1)%fact[n-l-1]+1, l+1, n);
		}
	}
	public static void sort(int[] array){
		ArrayList<Integer> copy = new ArrayList<Integer>();
		for (int i : array)
			copy.add(i);
		Collections.sort(copy);
		for(int i = 0;i<array.length;i++)
			array[i] = copy.get(i);
	}
	static String[] readArrayString(int n){
		String[] array = new String[n];
		for(int j =0 ;j<n;j++)
			array[j] = sc.next();
		return array;
	}
	static int[] readArrayInt(int n){
    	int[] array = new int[n];
    	for(int j = 0;j<n;j++)
    		array[j] = sc.nextInt();
    	return array;
    }
	static int[] readArrayInt1(int n){
		int[] array = new int[n+1];
		for(int j = 1;j<=n;j++){
			array[j] = sc.nextInt();
		}
		return array;
	}
	static long[] readArrayLong(int n){
		long[] array = new long[n];
		for(int j =0 ;j<n;j++)
			array[j] = sc.nextLong();
		return array;
	}
	static double[] readArrayDouble(int n){
		double[] array = new double[n];
		for(int j =0 ;j<n;j++)
			array[j] = sc.nextDouble();
		return array;
	}
	static int minIndex(int[] array){
		int minValue = Integer.MAX_VALUE;
		int minIndex = -1;
		for(int j = 0;j<array.length;j++){
			if (array[j] < minValue){
				minValue = array[j];
				minIndex = j;
			}
		}
		return minIndex;
	}
	static int minIndex(long[] array){
		long minValue = Long.MAX_VALUE;
		int minIndex = -1;
		for(int j = 0;j<array.length;j++){
			if (array[j] < minValue){
				minValue = array[j];
				minIndex = j;
			}
		}
		return minIndex;
	}
	static int minIndex(double[] array){
		double minValue = Double.MAX_VALUE;
		int minIndex = -1;
		for(int j = 0;j<array.length;j++){
			if (array[j] < minValue){
				minValue = array[j];
				minIndex = j;
			}
		}
		return minIndex;
	}
	static long power(long x, long y){
		if (y == 0)
			return 1;
		if (y%2 == 1)
			return (x*power(x, y-1))%mod;
		return power((x*x)%mod, y/2)%mod;
	}
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;
        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } 
                catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
        double nextDouble() {
            return Double.parseDouble(next());
        }
        String nextLine() {
            String str = "";
            try{
                str = br.readLine();
            } 
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
        
    }	
}

//StringJoiner sj = new StringJoiner(" "); 
//sj.add(strings)
//sj.toString() gives string of those stuff w spaces or whatever that sequence is