Moving 3

Jungyu is trapped in a maze of size $(N+1)\times(M+1)$. The maze is divided into rooms of size $1\times1$. The top-left room is $(0, 0)$ and the bottom-right room is $(N, M)$.

Jungyu is currently at $(0, 0)$ and wants to move to $(N, M)$. If Jungyu is at $(r, c)$, he can move through a door to $(r+1, c)$ or $(r, c+1)$. He cannot leave the maze.

This time there is trash in the gaps between the doors. So when moving from $(r, c)$ to $(r+1, c)$ or from $(r, c)$ to $(r, c+1)$, he must take all the trash in the door gap. Moving from $(r, c)$ to $(r+1, c)$ takes $A_r$ pieces of trash, and moving to $(r, c+1)$ takes $B_c$ pieces of trash.

Find the minimum amount of trash Jungyu must take when he moves to $(N, M)$.

The first line gives the size of the maze, $N$ and $M$. ($1 \le N, M \le 500{,}000$)

The next line gives $N$ integers $A_i$. ($1 \le A_i \le 10^9$)

The next line gives $M$ integers $B_i$. ($1 \le B_i \le 10^9$)

On the first line, print the amount of trash Jungyu must take when he moves to $(N, M)$.