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 (1.15.1) (judge verdict: Accepted)

use std::io::Write;

#[allow(unused)]
macro_rules! debug {
    ($($format:tt)*) => (write!(std::io::stderr(), $($format)*).unwrap());
}
#[allow(unused)]
macro_rules! debugln {
    ($($format:tt)*) => (writeln!(std::io::stderr(), $($format)*).unwrap());
}

fn main() {
    let n: usize = {
        let mut s = String::new();
        std::io::stdin().read_line(&mut s).unwrap();
        let s = s.trim_right().to_owned();
        s.parse().unwrap()
    };

    let a: Vec<usize> = {
        let mut v: Vec<usize> = Vec::new();
        for _ in 0..n {
            let v_i = {
                let mut s = String::new();
                std::io::stdin().read_line(&mut s).unwrap();
                let s = s.trim_right().to_owned();
                s.parse().unwrap()
            };
            v.push(v_i)
        }
        v
    };

    if a[0] != 0 {
        println!("{}", -1);
        return;
    }

    let mut ans = 0;
    ans += a[n - 1];
    for i in (0..(n - 1)).rev() {
        if a[i] + 1 < a[i + 1] {
            println!("{}", -1);
            return;
        } else if a[i] + 1 == a[i + 1] {
        } else {
            ans += a[i];
        }
    }
    println!("{}", ans);
}