Problem:
Score : 700 points

Problem Statement

There is a sequence X of length N, where every element is initially 0. Let X_i denote the i-th element of X.

You are given a sequence A of length N. The i-th element of A is A_i. Determine if we can make X equal to A by repeating the operation below. If we can, find the minimum number of operations required.

Choose an integer i such that 1\leq i\leq N-1. Replace the value of X_{i+1} with the value of X_i plus 1.

Constraints

1 \leq N \leq 2 \times 10^5

0 \leq A_i \leq 10^9(1\leq i\leq N)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

If we can make X equal to A by repeating the operation, print the minimum number of operations required. If we cannot, print -1.

Sample Input 1

4
0
1
1
2

Sample Output 1

3

We can make X equal to A as follows:

Choose i=2. X becomes (0,0,1,0).

Choose i=1. X becomes (0,1,1,0).

Choose i=3. X becomes (0,1,1,2).

Sample Input 2

3
1
2
1

Sample Output 2

-1

Sample Input 3

9
0
1
1
0
1
2
2
1
2

Sample Output 3

8

Solution in Rust (1.42.0) (judge verdict: Wrong Answer)

use proconio::input;

fn main() {
    input! {
        n: usize,
        aa: [usize; n],
    }

    if aa[0] != 0 {
        println!("-1");
        return;
    }
    for i in 0..n-1 {
        if aa[i+1] > aa[i] + 1 {
            println!("-1");
            return;
        }
    }

    let mut xx = vec![0; n];
    let mut ans = 0;

    loop {
        let mut index_max = None;
        let mut temp = 0;
        for i in (0..n).rev() {
            if aa[i] != xx[i] && temp < aa[i] {
                temp = aa[i];
                index_max = Some(i);
            }
        }
        if let Some(index) = index_max {
            for i in 1..=aa[index] {
                ans += 1;
                xx[index+i-aa[index]] = i;
            }
        } else {
            break;
        }
    }
    println!("{}", ans);
}