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A prime number (or a prime) is a natural number greater than 11 that cannot be formed by multiplying two smaller natural numbers.

Now lets define a number NN as the supreme number if and only if each number made up of an non-empty subsequence of all the numeric digits of NN must be either a prime number or 11.

For example, 1717 is a supreme number because 11, 77, 1717 are all prime numbers or 11, and 1919 is not, because 99 is not a prime number.

Now you are given an integer N\ (2 \leq N \leq 10^{100})N (2≤N≤10100), could you find the maximal supreme number that does not exceed NN?

Input

In the first line, there is an integer T\ (T \leq 100000)T (T≤100000) indicating the numbers of test cases.

In the following TT lines, there is an integer N\ (2 \leq N \leq 10^{100})N (2≤N≤10100).

Output

For each test case print "Case #x: y", in which xx is the order number of the test case and yy is the answer.

樣例輸入複製

2
6
100

樣例輸出複製

Case #1: 5
Case #2: 73

題目來源

題意：

要找一個不大於n的數 這個數的子序列都是質數或1

思路：

注意， 子序列是先後順序不變但是不一定連續。子串要求要連續。

所以這些數是有限的，而且個數很少，可以手算的出來

首先一位的只有1,2,3,5,7

2和5只能在最高位

除了1之外其他數只可能最多出現一次

因此，只有20個數：1, 2, 3, 5, 7, 11, 13, 17, 23, 31,37, 53, 71, 73, 113, 131, 137, 173, 311, 317

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<vector>
#include<cmath>
#include<set>
//#include<bits/stdc++.h>
#define inf 0x7f7f7f7f7f7f7f7f
using namespace std;
typedef long long LL;

int t;
char s[105];
int nums[100] = {1, 2, 3, 5, 7, 11, 13, 17, 23, 31,
 37, 53, 71, 73, 113, 131, 137, 173, 311, 317};


int main()
{
	cin >> t;
	for (int cas = 1; cas <= t; cas++) {
		cin >> s;
		if(strlen(s) >= 4){
            printf("Case #%d: 317\n", cas);
            continue;
		}
		int n = 0;
		for(int i = 0; i < strlen(s); i++){
            n *= 10;
            n += s[i] - '0';
		}

		int ans = 0;
        for(int i = 0; i < 20; i++){
            if(n >= nums[i]){
                ans = nums[i];
            }
            else{
                break;
            }
        }

        printf("Case #%d: %d\n", cas, ans);
	}
	return 0;
}