CF1702A.Round Down the Price

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题目描述

At the store, the salespeople want to make all prices round.

In this problem, a number that is a power of 1010 is called a round number. For example, the numbers 100=110^0 = 1 , 101=1010^1 = 10 , 102=10010^2 = 100 are round numbers, but 2020 , 110110 and 256256 are not round numbers.

So, if an item is worth mm bourles (the value of the item is not greater than 10910^9 ), the sellers want to change its value to the nearest round number that is not greater than mm . They ask you: by how many bourles should you decrease the value of the item to make it worth exactly 10k10^k bourles, where the value of kk — is the maximum possible ( kk — any non-negative integer).

For example, let the item have a value of 178178 -bourles. Then the new price of the item will be 100100 , and the answer will be 178100=78178-100=78 .

输入格式

The first line of input data contains a single integer tt ( 1t1041 \le t \le 10^4 ) — the number of test cases .

Each test case is a string containing a single integer mm ( 1m1091 \le m \le 10^9 ) — the price of the item.

输出格式

For each test case, output on a separate line a single integer dd ( 0d<m0 \le d < m ) such that if you reduce the cost of the item by dd bourles, the cost of the item will be the maximal possible round number. More formally: md=10km - d = 10^k , where kk — the maximum possible non-negative integer.

输入输出样例

  • 输入#1

    7
    1
    2
    178
    20
    999999999
    9000
    987654321

    输出#1

    0
    1
    78
    10
    899999999
    8000
    887654321

说明/提示

In the example:

  • 10=1001 - 0 = 10^0 ,
  • 21=1002 - 1 = 10^0 ,
  • 17878=102178 - 78 = 10^2 ,
  • 2010=10120 - 10 = 10^1 ,
  • 999999999899999999=108999999999 - 899999999 = 10^8 ,
  • 90008000=1039000 - 8000 = 10^3 ,
  • 987654321887654321=108987654321 - 887654321 = 10^8 .

Note that in each test case, we get the maximum possible round number.

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