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: Wrong Answer)

fun main(args: Array<String>) {
    val N = readInputLine().toInt()
    val arr = LongArray(N) { readInputLine().toLong() }
    val tmp = LongArray(N)
    
    if (arr[0] != 0L) {
        println(-1)
        return
    }

    var ans = 0L

    for (i in N - 1 downTo 1) {
        if (tmp[i] > arr[i]) {
            println(-1)
            return
        } else if (tmp[i] == arr[i]) {
            continue
        } else {
            if (arr[i] == 1L) {
                tmp[i] = 1L
                ans++
            } else {
                tmp[i] = arr[i]
                tmp[i - 1] = arr[i] - 1L
                ans += 2L
            }
        }
    }
    
    if (tmp[0] != 0L) {
        println(-1)
    } else {
        println(ans)
    }
}

fun readInputLine(): String {
    return readLine()!!
}