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: Time Limit Exceeded)

def solve0(N, A):
    p = -1
    for i, x in enumerate(A):
        if x > p + 1:
            return -1
        p = x
    r = 0
    p = set([0])
    for i in range(N - 1, -1, -1):
        c = 0
        q = set(n - 1 for n in p if n - 1 >= 0)
        q.add(A[i])
        r += len(q)
        if 0 in q:
            r -= 1
        print("*", A[i], c, r, p, q)
        p = q
    return r

def solve(N, A):
    p = -1
    for x in A:
        if x > p + 1:
            return -1
        p = x
    r = 0
    p = [0]
    for i in range(N - 1, -1, -1):
        q = [n - 1 for n in p if n >= 2]
        a = A[i]
        if a > 0 and (not q or q[-1] != a):
            q.append(a)
        r += len(q)
        # print("*", A[i], c, r, p, q)
        p = q
    return r

N = int(input())
A = [int(input()) for _ in range(N)]

print(solve(N, A))