//一次性全部生成完，想用多少用多少，时间长点
#include <iostream>#include<stdio.h>char min(char a, char b){ return (a>b)?b:a;}char max(char a, char b){ return (a>b)?a:b;}void swapIntOrChar(int a, int b){ a^=b^=a^=b;}bool isPrime(int n){    if(n==1) return false;    for(int i = 2; i*i <= n; i++) if(n % i == 0) return false;    return true;}int prime[78500];//装78498个素数,0号废弃。void generatePrimeArray(int n){ for(int i = 2, j = 1; i <= n; i++) if(isPrime(i)) prime[j++] = i;}/*-------- fibonacci -----*/unsigned long long f[50];//f[92]= 12200 16041 51218 76738 (20位),//unsigned long long最大18446744073709551615  //20位void generateFbnc( int n){//形成n个fibonacci数列数，供用。。    f[0] = f[1] = 1;    for(int i = 2; i < n; i++) f[i] = f[i-1] + f[i-2];}using namespace std;int main(){    int a, b, c, d, e;    //printf("%d\n", isPrime(1000003));    generatePrimeArray(1000003);    //printf("%d    %d\n", prime[78498], prime[78499]);    int loop_rounds;    while(~scanf("%d", &loop_rounds)){//输出最近的和距离。        while(loop_rounds--){            int t, record;            scanf("%d", &t);            if(isPrime(t)){                printf("%d 0\n", t);                continue;            }else if(t == 1){                printf("2 1\n");                continue;            }            /*找到第一个超过prime[i]的i*/            for(int i = 1; i <= 78499; i++){                if(t < prime[i]){                    record = i;                    break;                }            }            record -= 1;            //三者成等差.            if(prime[record] + prime[record+1] == 2* t){                printf("%d %d\n", prime[record], t-prime[record]);                continue;            }            if(t-prime[record] < prime[record+1]-t) printf("%d %d\n", prime[record], t-prime[record]);            else printf("%d %d\n", prime[record+1], prime[record+1]-t);        }    }    return 0;}/*样例输入36810样例输出5 17 111 1*/