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A15227. Round Down the Price

编程题 普及/提高-

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 $10$ is called a round number. For example, the numbers $10^0 = 1$ , $10^1 = 10$ , $10^2 = 100$ are round numbers, but $20$ , $110$ and $256$ are not round numbers.

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

For example, let the item have a value of $178$ -bourles. Then the new price of the item will be $100$ , and the answer will be $178-100=78$ .

输入格式

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

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

输出格式

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

- $1 - 0 = 10^0$ ,
- $2 - 1 = 10^0$ ,
- $178 - 78 = 10^2$ ,
- $20 - 10 = 10^1$ ,
- $999999999 - 899999999 = 10^8$ ,
- $9000 - 8000 = 10^3$ ,
- $987654321 - 887654321 = 10^8$ .

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