Diagonal sums

You are given an $n \times n$ square matrix. Find the sum of the main diagonal (from the top-left to the bottom-right corner) and of the anti-diagonal (from the top-right to the bottom-left corner).

The first line contains $n$ ($1 \le n \le 100$). Each of the next $n$ lines contains $n$ integers ($|a_{ij}| \le 10^9$).

A single line with two numbers: the main diagonal sum, then the anti-diagonal sum.

On the main diagonal $i = j$, on the anti-diagonal $i + j = n - 1$ (counting from $0$). Two nested loops are not needed — one loop is enough. In the first example: $1 + 5 + 9 = 15$ and $3 + 5 + 7 = 15$. Sums reach $10^{11}$ — long long.