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.util.*;
import java.util.concurrent.ThreadLocalRandom;
import java.io.*;

public class Multiple {

	public static void main(String[] args) throws NumberFormatException, IOException, InterruptedException {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		int n=sc.nextInt();
		long[]arr=new long[n];
		for (int i = 0; i < arr.length; i++) {
			arr[i]=sc.nextLong();
		}
		pw.println("1 1");
		pw.println(-1*arr[0]);
		arr[0]=0;
		if(n==1) {
			pw.println("1 1");
			pw.println("0");
			pw.println("1 1");
			pw.println("0");
			pw.close();
			return;
		}
		pw.println("2 "+arr.length);
		StringBuilder sb=new StringBuilder();
		for (int i = 1; i < arr.length; i++) {
			sb.append(1l*arr[i]*(arr.length-1)+" ");
			arr[i]+=1l*arr[i]*(arr.length-1);
		}
		pw.println(sb.substring(0,sb.length()-1));
		pw.println("1 "+arr.length);
		sb=new StringBuilder();
		for (int i = 0; i < arr.length; i++) {
			sb.append(-1*arr[i]+" ");
		}
		pw.println(sb.substring(0,sb.length()-1));
		pw.close();
	}

	static void shuffleArray(int[] ar) {
		// If running on Java 6 or older, use `new Random()` on RHS here
		Random rnd = ThreadLocalRandom.current();
		for (int i = ar.length - 1; i > 0; i--) {
			int index = rnd.nextInt(i + 1);
			// Simple swap
			int a = ar[index];
			ar[index] = ar[i];
			ar[i] = a;
		}
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public Scanner(FileReader r) {
			br = new BufferedReader(r);
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public int[] nextIntArr(int n) throws IOException {
			int[] arr = new int[n];
			for (int i = 0; i < arr.length; i++)
				arr[i] = nextInt();
			return arr;
		}

		public boolean ready() throws IOException {
			return br.ready();
		}
	}

}