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 PYTHON3):

Solution:
n=int(input())
a=list(map(int,input().split()))

if n!=1:
    print(n,n)
    print(-a[-1])


#in short its like ...suppose a[i] to be x
#first chance is wasted on any element of length 1
#so after second chance...x will beacome x+ (n-1)*x that is n*x
#in third step we will take all element so length n and we will do
#x = x_after_second_step +(-n*x)
#where x was initially n*x....so it will be x = n*x-n*x
#that is zero..

#and in this iteration of lemgth n..the last element that is already zero..we will 
#add zero*n to it that will be zero only...

#hhahhhahaha


    print(1,n-1)
    res_2 = [a[i]*(n-1) for i in range(n-1)]
    print(*res_2)

    print(1,n)
    res_3 = [-a[i]*n for i in range(n)]
    res_3[-1] = 0

    print(*res_3)
else:
    print(1,1)
    print(-a[0])
    #to waste second chance
    print(1,1)
    print(0)
    #to waste third chance
    print(1,1)
    print(0)