time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an integer x𝑥. Your task is to find any integer y𝑦 (1≤y<x)(1≤𝑦<𝑥) such that gcd(x,y)+ygcd(𝑥,𝑦)+𝑦 is maximum possible.

Note that if there is more than one y𝑦 which satisfies the statement, you are allowed to find any.

gcd(a,b)gcd(𝑎,𝑏) is the Greatest Common Divisor of a𝑎 and b𝑏. For example, gcd(6,4)=2gcd(6,4)=2.

Input

The first line contains a single integer t𝑡 (1≤t≤10001≤𝑡≤1000) — the number of test cases.

Each of the following t𝑡 lines contains a single integer x𝑥 (2≤x≤10002≤𝑥≤1000).

Output

For each test case, output any y𝑦 (1≤y<x1≤𝑦<𝑥), which satisfies the statement.

Example

input

Copy

7

10

7

21

100

2

1000

6

output

Copy

5
6
18
98
1
750
3

解题说明：此题是一道数学题，分析后能知道最大肯定不会超过x，可以采用贪心算法，y选择为x-1，这样gcd(x,y)为0，则最大值就是x-1

#include <stdio.h>int main()
{int t = 0, n = 0;scanf("%d", &t);while (t--){scanf("%d", &n);printf("%d\n", n - 1);}return 0;
}