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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
/**
 * @author Lin
 * @version 1.0
 * @date 2020/12/12 0:37
 */
public class MultiplesOfLength
{
	public static void main(String[] args) throws IOException
	{
		
		StringBuffer stringBuffer = new StringBuffer();
		int n = Reader.nextInt();
		long ans = 0;
		//两行数组输入
		int[] an = new int[n];
		for (int i = 0; i < n; i++)
		{
			an[i] = Reader.nextInt();
		}
		if (n == 1)
		{
			stringBuffer.append("1 " + n + "\n");

			stringBuffer.append(-(long) an[0] * n + " ");

			stringBuffer.append("\n1 "+1 + "\n");
			stringBuffer.append((long) 0 + " ");
			stringBuffer.append("\n1 " + 1 + " \n");
			stringBuffer.append(0 + " \n");
		}else
		{
			stringBuffer.append("1 " + n + "\n");
			for (int j = 0; j < n; j++)
			{
				stringBuffer.append(-(long) an[j] * n + " ");
			}
			stringBuffer.append("\n1 "+(n - 1) + "\n");
			for (int j = 0; j < n - 1; j++)
			{
				stringBuffer.append((long) an[j] * (n - 1) + " ");
			}
			stringBuffer.append("\n2 " + n + " \n");
			for (int i = 1; i < n - 1; i++)
			{
				stringBuffer.append("0 ");
			}
			stringBuffer.append((long)an[n - 1]*(n-1) + " \n");
		}

		System.out.println(stringBuffer.toString());
	}
	static class Reader {
		static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		static StringTokenizer tokenizer = new StringTokenizer("");
		static String nextLine() throws IOException
		{// 读取下一行字符串
			return reader.readLine();
		}
		static String next() throws IOException {// 读取下一个字符串
			while (!tokenizer.hasMoreTokens()) {
				tokenizer = new StringTokenizer(reader.readLine());
			}
			return tokenizer.nextToken();
		}

		static int nextInt() throws IOException {// 读取下一个int型数值
			return Integer.parseInt(next());
		}
		static long nextLong() throws IOException {// 读取下一个int型数值
			return Long.parseLong(next());
		}
		static double nextDouble() throws IOException {// 读取下一个double型数值
			return Double.parseDouble(next());
		}
	}
}