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;
#define PII pair<int,int>
#define X first
#define Y second
#define PB push_back
#define SORT(a) sort(a.begin(), a.end())
#define CLR(a) memset(a, 0, sizeof(a))
#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define FOE(i,a,b) for (int i=(a);i<=(b);i++)
#define FOREACH(i,a) for (auto i = (a).begin(); i != (a).end(); i++)
#define TRA(i,a,b) for (int i = (a); i != -1; i = (b)[i])
#define INF (1 << 30)
#define EPS (1e-9)
#define REP(i,n) FOR(i,0,n)
#define LL long long
#define MOD 1000000007
#define N 2001010
int n, m;
int a[N] = {0};
int main(){
        ios::sync_with_stdio(false);
        cin >> n;
        FOR(i,0,n) cin >> a[i];
        LL ans = 0;
        for (int i = n - 1; i >= 0; i--){
                if (a[i] > i){
                        cout << "-1\n";
                        return 0;
                }
                if (a[i] != a[i + 1] - 1)
                        ans += a[i];
        }
        cout << ans << '\n';
        return 0;
}