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;
use std::cmp;
use std::collections::BinaryHeap;
use std::collections::HashMap;
use std::collections::HashSet;
use std::f64::consts::PI;

fn read_line() -> String {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).ok();
    buf
}

fn func(a: &Vec<usize>) -> i64 {
    let n = a.len() - 1;
    let mut ans = 0;
    for i in (0..n).rev() {
        if a[i+1] == a[i] + 1 {
            continue;
        } else if a[i] <= i && a[i] + 1 > a[i+1] {
            ans += a[i] as i64;
        } else {
            return -1;
        }
    }
    ans
}


fn main() {
    let n: usize = read_line().trim().parse().unwrap();
    let mut a = Vec::with_capacity(n+1);
    for i in 0..n {
        let x: usize = read_line().trim().parse().unwrap();
        a.push(x);
    }
    a.push(0);
    println!("{}", func(&a));
}