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 <stdio.h>
#include <vector>
#include <queue>
#include <algorithm>
#include <iostream>
#include <string>
#include <bitset>
#include <map>
#include <set>
#include <tuple>
#include <string.h>
#include <math.h>
#include <random>
#include <functional>
#include <assert.h>
#include <math.h>
#include <sstream>
#include <iterator>

using namespace std;

using i64 = long long int;
using ii = pair<int, int>;
using ii64 = pair<i64, i64>;

int main()
{
    int n;
    scanf("%d", &n);

    vector<i64> arr(n);

    for (int i = 0; i < n; i++)
        scanf("%lld", &arr[i]);

    if (arr[0] != 0)
    {
        printf("-1\n");
        return 0;
    }

    i64 ans = 0;

    for (int i = 0; i < n - 1; i++)
    {
        if (arr[i + 1] > arr[i] + 1)
        {
            printf("-1\n");
            return 0;
        }

        if (arr[i + 1] <= arr[i])
            ans += arr[i];
    }

    ans += arr[n - 1];

    printf("%lld\n", ans);

    return 0;
}