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 C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
int main(void)
{
    long int n;
    scanf("%ld\n",&n);
    long long int a[200005]={};
    for(int p=0;p<n;p++)scanf("%lld\n",&a[p]);
    long int pointer=n-1;
    long long int sum=0;
    long long int comp[200005]={};
    while(pointer>=0)
    {
        if(comp[pointer]<a[pointer])
        {
            for(int q=pointer;q>pointer-a[pointer];q--)
            {
                comp[q]=a[pointer]-pointer+q;
                if(comp[q]>q || comp[q]>a[q])
                {
                    printf("-1\n");
                    return 0;
                }
                sum++;
            }
        }
        pointer--;
    }
    printf("%lld",sum);
    return 0;
}