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: Time Limit Exceeded)

#include <iostream>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);++i)
#define rrep(i,n) for(int i=(n);i>=0;--i)
#define out(v) cout<<v<<"\n"

int main() {
  cin.tie(0); ios::sync_with_stdio(false);
  int N; cin >> N;

  int A[N], X[N];
  rep(i, N) {
    cin >> A[i];
    X[i] = 0;
  }
  int ans = 0;
  rrep(i, N-1) {
    if (A[i] > 0) {
      if (i - A[i] < 0) {
        out(-1);
        return 0;
      }
      rep(j, A[i] + 1) {
        if (A[i-j] < A[i] - j) {
          out(-1);
          return 0;
        }
        if (X[i-j] != A[i] - j) {
          X[i-j] = A[i] - j;
          ++ans;
        }
      }
    }
  }
  out(ans);

  return 0;
}