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 (3.4.3) (judge verdict: Runtime Error)

import math
n = int(input())
lis = []
for i in range(n):lis.append(int(input()))
if lis[0] != 0:
    print("-1")
    exit()
cou = 0
print(lis,cou)
for k in range(n-1):
    if lis[n-(k+1)] > lis[n-(k+2)]:
        a =  math.ceil(lis[n-(k+1)] / (lis[n-(k+2)]+1))
        cou += a
        lis[n-(k+1)] = 0
        lis[n-(k+2)] = min(lis[n-(k+1)]-1,lis[n-(k+2)],math.ceil(lis[n-(k+1)]/a))
    else:
        cou += lis[n-(k+1)]
        lis[n-(k+2)] -= lis[n-(k+1)]-1
        lis[n-(k+1)] = 0
    print(lis,cou)
print(cou)