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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pii;
typedef vector<pii> vii;

int a[200000];

int main() {

    int n, i; ll ans = 0;

    scanf("%d", &n);
    for(i = 0; i < n; i++) scanf("%d", a + i);
    for(i = 1; i < n && a[i] - a[i - 1] <= 1; i++);
    if (a[0] || i < n) {
        puts("-1");
        return 0;
    }
    for(i = n - 1; i >= 0; i--) {
        if (!a[i]) continue;
        if (i < n - 1 && a[i + 1] == a[i] + 1) continue;
        ans += a[i];
    }
    printf("%lld\n", ans);

    return 0;
}