nums.mod to 10^9 + 7 for taking modulo.sumWidths to 0, which will store the final result.i.
i and for n - i - 1 (where n is the length of nums), taking modulo mod.sumWidths the contribution of nums[i] as a maximum, which is nums[i] * (2^i - 1).sumWidths the contribution of nums[i] as a minimum, which is nums[i] * (2^(n - i - 1) - 1).mod after each addition and subtraction.sumWidths as the final answer.