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: Accepted)

#![allow(unused, non_snake_case, dead_code, non_upper_case_globals)]
use {
    proconio::{marker::*, *},
    std::{cmp::*, collections::*, mem::*},
};
fn main() {
    input! {//
        n:usize,
        a:[i64;n],
    }
    if a[0] != 0 {
        println!("{}", -1);
        return;
    }
    let mut ans = 0i64;
    for i in 0..n {
        if i < n - 1 && a[i + 1] - a[i] > 1 {
            println!("{}", -1);
            return;
        }
        if i >= n - 1 || a[i] + 1 != a[i + 1] {
            ans += a[i];
        }
    }
    println!("{}", ans);
}