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 Python (2.7.6) (judge verdict: Wrong Answer)

#!/bin/python

import os
import sys

#
# Complete the gradingStudents function below.
#
def solution(arr):
    if arr[0] != 0:
        return -1
    sum = 0
    for i in range(1,len(arr)):
        if (i>1) and (arr[i] > 1) and (arr[i-1] == 0) and (arr[i-2] == 0):
            return -1
        if (arr[i-1] >= arr[i]) and (arr[i-1] != 0):
            sum += 1
        elif (arr[i-1] < arr[i]):
            sum += arr[i]-arr[i-1]
    return sum
            

if __name__ == '__main__':
    
    n = int(raw_input())
    arr = []
    for i in range(n):
        arr.append(int(raw_input()))

    result = solution(arr)
    print str(result)