Untitled

public class UserMainCode {
    // Function to count the number of possible ways to set up the colonies.
    public static int Colonies(int input1, int input2) {
        // Since ants can only move up, down, left and right, each cell in the grid can only
        // be part of a colony if it's adjacent (not diagonally) to another cell that's part of a colony.
        // This is essentially a combinatorial problem where we need to count the number of ways to fill
        // a grid with blocks such that each block is connected to at least one other block horizontally
        // or vertically.
        
        // A dynamic programming approach can be used where dp[i][j] represents the number of ways to arrange
        // colonies in a grid of size i x j. 
        // The base case would be when either i or j is 1, because the colonies can only be set up in a linear
        // fashion, thus the number of ways to set up the colonies is 2^(i*j) - 1 (all combinations of cells being
        // filled or not, minus the case where no cell is filled).
        
        // Initialize a 2D array for dynamic programming
        long[][] dp = new long[input1 + 1][input2 + 1];
        
        // Modulo value for large numbers
        long MOD = 1000000007;
        
        // Base cases
        for (int i = 1; i <= input1; i++) {
            for (int j = 1; j <= input2; j++) {
                if (i == 1 || j == 1) {
                    dp[i][j] = (pow(2, i * j) - 1 + MOD) % MOD; // Using pow function with modulo
                } else {
                    // Apply the recursive relation
                    // dp[i][j] is the sum of ways of arranging colonies in the smaller grid plus the number of ways
                    // to extend the grid with an additional row or column of colonies.
                    // Since the problem does not specify a pattern for the arrangement of colonies, we'll assume the
                    // simplest case where the arrangement of colonies in a smaller grid can be extended by adding a row
                    // or column of colonies in all possible ways.
                    dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]) % MOD;
                }
            }
        }
        
        // Return the number of ways for the original grid size.
        return (int)dp[input1][input2];
    }
    
    // Function to compute the power of a number with modulo
    public static long pow(long base, long exponent) {
        long result = 1;
        long MOD = 1000000007;
        
        while (exponent > 0) {
            if (exponent % 2 == 1) {
                result = (result * base) % MOD;
            }
            base = (base * base) % MOD;
            exponent = exponent / 2;
        }
        
        return result;
    }
}