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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

using ll = long long int;

const int MAX_N = (int)(2 * 1e5 + 5);

int n;
ll a[MAX_N], b[MAX_N];

int ans;

int main(void){
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%lld", &(a[i]));
  }

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

  for (int i = 1; i < n; ++i) {
    if (a[i] - a[i - 1] > 1 || a[i] < a[i - 1]) {
      printf("%d\n", -1);
      return 0;
    }
  }

  for (int i = 1; i < n; ++i) {
    if (a[i] == a[i - 1] + 1) {
      ans += 1;
    }
    else {
      ans += a[i];
    }
  }

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

  return 0;
}