Problem: You are given an array a of n integers.

You want to make all elements of a equal to zero by doing the following operation exactly three times:

  * Select a segment, for each number in this segment we can add a multiple of len to it, where len is the length of this segment (added integers can be different). 



It can be proven that it is always possible to make all elements of a equal to zero.

Input

The first line contains one integer n (1 ≤ n ≤ 100 000): the number of elements of the array.

The second line contains n elements of an array a separated by spaces: a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9).

Output

The output should contain six lines representing three operations.

For each operation, print two lines:

  * The first line contains two integers l, r (1 ≤ l ≤ r ≤ n): the bounds of the selected segment.

  * The second line contains r-l+1 integers b_l, b_{l+1}, ..., b_r (-10^{18} ≤ b_i ≤ 10^{18}): the numbers to add to a_l, a_{l+1}, …, a_r, respectively; b_i should be divisible by r - l + 1.

Example

Input


4
1 3 2 4


Output


1 1 
-1
3 4
4 2
2 4
-3 -6 -6

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class A {

	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		int n = sc.nextInt();
		long[] arr = new long[n];
		long[] g = gcdex(n, n-1);
		for(int i = 0; i < n; i++){
			arr[i] = sc.nextInt();
		}
		if(n == 1) {
			System.out.println("1 1");
			System.out.println(-arr[0]);
			System.out.println("1 1");
			System.out.println("0");
			System.out.println("1 1");
			System.out.println("0");
		}
		else {
			StringBuilder sb = new StringBuilder();
			sb.append(1+" "+n+"\n");
			for(int i = 0; i < n; i++) {
				sb.append((-arr[i]*g[0]*n));
				if(i < n) sb.append(" ");
			}
			sb.append("\n"+2+" "+n+"\n");
			for(int i = 1; i < n; i++) {
				sb.append((-arr[i]*g[1]*(n-1)));
				if(i < n) sb.append(" ");
			}
			sb.append("\n"+1+" "+1+"\n"+(-arr[0]+arr[0]*g[0]*n));
			PrintWriter pw = new PrintWriter(System.out);
			pw.print(sb.toString());
			pw.flush();
		}
	}
	
	static long[] gcdex(long a, long b){
		if(b > a) {
			long[] p = gcdex(b, a);
			return new long[] {p[1], p[0]};
		}
		else if(b == 0) return new long[] {1, 0};
		else{
			long[] p = gcdex(b, a % b);
			return new long[] {p[1], p[0] - p[1]*(a/b)};
		}
	}
	
	static class FastScanner {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
        public FastScanner() {
            reader = new BufferedReader(new InputStreamReader(System.in), 32768);
            tokenizer = null;
        }
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong() {
        	return Long.parseLong(next());
        }
        public double nextDouble() {
        	return Double.parseDouble(next());
        }
        public String nextLine() {
        	try {
        		return reader.readLine();
        	} catch(IOException e) {
        		throw new RuntimeException(e);
        	}
        }
    }
	

}