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 <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <utility>
#include <cstdio>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <cmath>
#include <set>
#include <map>
#define my_abs(x) ((x) < 0 ? -(x) : (x))
#define mp std::make_pair
#define pb push_back
typedef long long ll; 
int arr[200005]; 
int main()
{
	// freopen("AGC024-C.in", "r", stdin); 
	int n; 
	scanf("%d", &n); 
	for (int i = 0; i < n; i++)
		scanf("%d", arr + i); 
	int cur = n;
	long long ans = 0; 
	for (int i = n - 1; i >= 0; i--)
	{
		if (i - arr[i] + 1 == cur)
			continue; 
		if (i - arr[i] + 1 >= cur)
		{
			puts("-1"); 
			return 0; 
		}
		ans += arr[i]; 
		cur = i - arr[i] + 1; 
		if (cur <= 0)
		{
			puts("-1"); 
			return 0; 
		}
	}
	printf("%lld\n", ans);
	return 0; 
}