2140B. Another Divisibility Problem

constructive algorithms, math, number theory

900, https://codeforces.com/problemset/problem/2140/B

Alice and Bob are playing a game in which Alice has given Bob a positive integer 𝑥<108.

To win the game, Bob has to find another positive integer 𝑦<109 such that 𝑥#𝑦 is divisible by 𝑥+𝑦.

Here 𝑥#𝑦 denotes the integer formed by concatenating the integers 𝑥 and 𝑦 in that order. For example, if 𝑥=835, 𝑦=47, then 𝑥#𝑦=83547.

However, since Bob is dumb, he is unable to find such an integer. Please help him.

It can be shown that such an integer always exists.

Input

Each test contains multiple test cases. The first line contains the number of test cases 𝑡 (1≤𝑡≤104). The description of the test cases follows.

The only line of each test case contains a single integer 𝑥 (1≤𝑥<108) — the integer that Alice has given to Bob.

Output

For each test case, print a single integer 𝑦 (1≤𝑦<109) so that Bob can win the game.

If there are multiple answers, print any one of them.

Example

input

6
8
42
1000
66666
106344
9876543

output

1
12
998
7872
8190
174036

Note

For the first test case, 𝑥=8, we can choose 𝑦=1, and we have 𝑥#𝑦=81, which is divisible by 𝑥+𝑦=9.

For the second test case, 𝑥=42, we can choose 𝑦=12, and we have 𝑥#𝑦=4212, which is divisible by 𝑥+𝑦=54.

t = int(input())
for _ in range(t):
    x = int(input())
    print(2 * x)

正确：因为 y = 2x 时，x#y = x * 10^d + 2x，x + y = 3x，而 (10^d + 2) 总是能被 3 整除。

高效：时间复杂度 O(1) 每个测试用例。

数学思路：

• 对于一个n位的数字x和一个m位的数字y，x#y=x*10^m+y。
• 当考虑x+y是否能整除x*10^m+y时，这等价于整除x*(10^m-1)`。
• 由于x+y显然不会直接整除x，因此需要让x+y整除(10^m-1)，即9*1111...(m位)。
• 为了满足上述条件，可以设定x+y=10^m-1，这意味着y的值可以通过公式y=10^(n+1)-1-x计算得出，其中n是x的位数。
• y的位数必须大于等于x的位数，才能保证了x+y>10^m-1。

写成 y = 10^(n+1)-1-x即可

for _ in range(int(input())):
    x = int(input())
    n = len(str(x))
    print(10**(n+1)-1-x)