C++ Easy Solution
\n#define ll long long int\nclass Solution {\npublic:\n    int maxArea(int h, int w, vector<int>& horizontalCuts,   vector<int>& verticalCuts) {\n        //addi
```\n#define ll long long int\nclass Solution {\npublic:\n    int maxArea(int h, int w, vector<int>& horizontalCuts,   vector<int>& verticalCuts) {\n        //adding h to the Horizontal Array\n        horizontalCuts.push_back(0);\n        horizontalCuts.push_back(h);\n        \n        //adding w to the Vertical Array\n        verticalCuts.push_back(0);\n        verticalCuts.push_back(w);\n        \n        //sorting both the array\n        sort(horizontalCuts.begin(), horizontalCuts.end());\n        sort(verticalCuts.begin(), verticalCuts.end());\n        \n        ll max_length = 0, max_breadth = 0;\n        \n        //finding the maximum consecutive distance by providing maximum Length to the Vertical Cut array and the maximum breadth to the horizontal cut array.\n        for(int i=1; i<horizontalCuts.size(); i++){\n            if(max_breadth < (horizontalCuts[i] - horizontalCuts[i-1])){\n                max_breadth = horizontalCuts[i] - horizontalCuts[i-1];\n            }\n        }\n        \n        for(int i=1; i<verticalCuts.size(); i++){\n            if(max_length < (verticalCuts[i] - verticalCuts[i-1])){\n                max_length = verticalCuts[i] - verticalCuts[i-1];\n            }\n        }\n        \n        //maximum area and returning by MODULUS OF 10^9+7\n        ll ans = (max_length * max_breadth);\n        return (ans)%1000000007;\n    }\n};\n```
['C']