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 <bits/stdc++.h>
typedef long long int ll;
typedef long double ld;
#define pb push_back
#define pii pair < int, int >
#define F first
#define S second
#define int  long long int
#define sync ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#pragma GCC optimize ("Ofast")
#pragma GCC optimize ("unroll-loops")
using namespace std;
const int N=2e5+100;
ll a[N];
int32_t main(){
    ll n;
    cin >> n;
    for (int i=1;i<=n;i++){
        cin >> a[i];
    }
    if (a[1]!=0){
        cout << -1 << endl;
        return 0;
    }
    ll ans=0;
    for (int i=2;i<=n;i++){
        if (a[i]-a[i-1]>1){
            cout << -1 << endl;
            return 0;
        }
        if (a[i]-a[i-1]==1){
            ans++;
        }
        else{
            ans+=a[i];
        }
    }
    cout << ans;
}