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: Time Limit Exceeded)

import heapq


class Value:
    def __init__(self, val, f):
        self.val = val
        self.key = f(val)

    def __lt__(self, other):
        return self.key < other.key

    def __repr__(self):
        return repr(self.val)


class Heapq:
    def __init__(self, arr, f=lambda x: x):
        for i in range(len(arr)):
            arr[i] = Value(arr[i], f)
        self.func = f
        self.hq = arr
        heapq.heapify(self.hq)

    def pop(self):
        return heapq.heappop(self.hq).val

    def push(self, a):
        heapq.heappush(self.hq, Value(a, self.func))

    def top(self):
        return self.hq[0].val


n = int(input())
a = [int(input()) for _ in range(n)]
if a[0] > 0:
    print(-1)
    exit(0)
hq = Heapq([], f=lambda x: (-a[x], x))
for i, x in enumerate(a):
    if x > 0:
        hq.push(i)
ans = 0
while hq.hq:
    i = hq.pop()
    if a[i] - 1 == a[i - 1]:
        ans += 1
        a[i] = (a[i + 1] - 1 if i != n - 1 else 0)
        if a[i] > 0:
            hq.push(i)
    else:
        print(-1)
        exit(0)
print(ans)