The operator for exponentiation is different from the addition, subtraction, multiplication or division operators in the sense that the de- fault associativity for exponentiation goes right to left instead of left to right. So unless we mess it up by placing parenthesis, 232 should mean 2(32) = 29 = 512 not (23)2 = 82 = 64 . This leads to the obvious fact that if we take the levels of exponents higher (i.e., 2ˆ3ˆ4ˆ5ˆ3), the numbers can become quite big. But let’s not make life miserable. We being the good guys would force the ultimate value to be no more than 10000. Givena1,a2,a3,...,aN andm(=10000)youonlyneedtocomputea1ˆa2ˆa3ˆ...ˆaN modm. Input There can be multiple (not more than 100) test cases. Each test case will be presented in a single line. The first line of each test case would contain the value for M (2 ≤ M ≤ 10000). The next number of that line would be N (1 ≤ N ≤ 10). Then N numbers — the values for a1,a2,a3,...,aN would follow. You can safely assume that 1 ≤ ai ≤ 1000. The end of input is marked by a line containing a single hash (‘#’) mark. Output For each of the test cases, print the test case number followed by the value of a1ˆa2ˆa3ˆ...ˆaN mod m on one line. The sample output shows the exact format for printing the test case number. Sample Input 10 4 2 3 4 5 100 2 5 2 53 3 2 3 2

Sample Output Case #1: 2 Case #2: 25 Case #3: 35