CF1242C.Sum Balance

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题目描述

Ujan has a lot of numbers in his boxes. He likes order and balance, so he decided to reorder the numbers.

There are kk boxes numbered from 11 to kk . The ii -th box contains nin_i integer numbers. The integers can be negative. All of the integers are distinct.

Ujan is lazy, so he will do the following reordering of the numbers exactly once. He will pick a single integer from each of the boxes, kk integers in total. Then he will insert the chosen numbers — one integer in each of the boxes, so that the number of integers in each box is the same as in the beginning. Note that he may also insert an integer he picked from a box back into the same box.

Ujan will be happy if the sum of the integers in each box is the same. Can he achieve this and make the boxes perfectly balanced, like all things should be?

输入格式

The first line contains a single integer kk ( 1k151 \leq k \leq 15 ), the number of boxes.

The ii -th of the next kk lines first contains a single integer nin_i ( 1ni50001 \leq n_i \leq 5\,000 ), the number of integers in box ii . Then the same line contains nin_i integers ai,1,,ai,nia_{i,1}, \ldots, a_{i,n_i} ( ai,j109|a_{i,j}| \leq 10^9 ), the integers in the ii -th box.

It is guaranteed that all ai,ja_{i,j} are distinct.

输出格式

If Ujan cannot achieve his goal, output "No" in a single line. Otherwise in the first line output "Yes", and then output kk lines. The ii -th of these lines should contain two integers cic_i and pip_i . This means that Ujan should pick the integer cic_i from the ii -th box and place it in the pip_i -th box afterwards.

If there are multiple solutions, output any of those.

You can print each letter in any case (upper or lower).

输入输出样例

  • 输入#1

    4
    3 1 7 4
    2 3 2
    2 8 5
    1 10
    

    输出#1

    Yes
    7 2
    2 3
    5 1
    10 4
    
  • 输入#2

    2
    2 3 -2
    2 -1 5
    

    输出#2

    No
    
  • 输入#3

    2
    2 -10 10
    2 0 -20
    

    输出#3

    Yes
    -10 2
    -20 1
    

说明/提示

In the first sample, Ujan can put the number 77 in the 22 nd box, the number 22 in the 33 rd box, the number 55 in the 11 st box and keep the number 1010 in the same 44 th box. Then the boxes will contain numbers {1,5,4}\{1,5,4\} , {3,7}\{3, 7\} , {8,2}\{8,2\} and {10}\{10\} . The sum in each box then is equal to 1010 .

In the second sample, it is not possible to pick and redistribute the numbers in the required way.

In the third sample, one can swap the numbers 20-20 and 10-10 , making the sum in each box equal to 10-10 .

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