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 PyPy3 (2.4.0) (judge verdict: Accepted)

import sys
input = sys.stdin.readline
N = int(input())
a = []
for _ in range(N): a.append(int(input()))
if a[0] != 0:
  print(-1)
  exit(0)
for i in range(N - 1):
  if a[i + 1] > a[i] and (a[i + 1] - a[i] > 1):
    print(-1)
    exit(0)
res = 0
s = 0
x = 1
for i in range(1, N):
  if a[i] == x:
    s += 1
  else:
    if x > 0: res += (s - 1) * x + 1
    if x > a[i] and a[i] > 1:
      res += a[i] - 1
    s = 1
    x = a[i]
  #print(i, res)
if x > 0: res += (s - 1) * x + 1
print(res)