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 Kotlin (1.0.0) (judge verdict: Accepted)

private fun readStrings() = readLine()!!.split(" ")
private fun readInts() = readStrings().map { it.toInt() }

fun main(args: Array<String>) {
    val N = readInts()[0]
    val a = (1 .. N).map { readInts()[0] }

    if (a[0] != 0) println(-1)
    else {
        var counts = 0L
        var prev = a[0]
        for (num in a.drop(1)) {
            if (num > prev) {
                if (num - prev > 1) {
                    println(-1)
                    return
                } else {
                    counts++
                }
            } else {
                counts += num
            }
            prev = num
        }
        println(counts)
    }
}