SDNU 1227.A great challenge. (Erichsen sieve)

Description

In the legend, there lived the most intelligent person named Haroro in SDNU. Haroro likes to challenge other people on the prime problem. She is an unbeaten myth. Now, turn to you. If you can find out the most close to its prime number faster than Haroro, you will receive an unexpected gift.

eg： If the screen appears 22, your answer should be 23;If the screen appears 8, your answer should be 7; if the X itself is prime, your answer should be X;if the nearest X prime number is two, then the answer is greater than its prime.

Input

The first line of the input contains the single number T(1 ≤ T≤ 100). Each of next T lines contains one number X(1≤X≤1000).

Output

The output contains N line, your answer should be the most close to its prime.

 

Sample Input

3
22
7
8

Sample Output

23
7
7

Hint

 

Source

Unknown

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int inf = 0x3f3f3f3f;
const int maxn = 1000 + 8;

int t, n, tot, pri[maxn];
bool is[maxn];

void e()
{
    tot = 0;
    memset(is, 1, sizeof(is));
    is[0] = is[1] = 0;
    for(int i = 2; i < maxn; i++)
    {
        if(is[i])
        {
            pri[++tot] = i;
            for(int j = i + i; j <= maxn; j += i)
                is[j] = 0;
        }
    }
}

int main()
{
//    std::ios::sync_with_stdio(0);
//    cin.tie(0);
//    cout.tie(0);
    e();
    scanf("%d", &t);
    for(int i = 0; i < t; i++)
    {
        scanf("%d", &n);
        int m = upper_bound(pri + 1, pri + tot + 1, n) - pri;
        if(abs(pri[m] - n) > abs(pri[m - 1] - n))
            printf("%d\n", pri[m - 1]);
        else
            printf("%d\n", pri[m]);
    }
    return 0;
}