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 REP(i,a,b) for(int i=a;i<b;i++)
#define rep(i,n) REP(i,0,n)

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<ll> vl;

int N;
ll ans = 0;
int f = 0;

void solve() {
    cin >> N;
    int pre;
    rep(i,N) {
        int A;
        cin >> A;
        if (A > i) {
            f = 1;
            continue;
        }
        if (i > 0) {
            if (A != pre+1) {
                ans += pre;
            }
        }
        pre = A;
    }
    ans += pre;
    if (f) cout << -1 << endl;
    else cout << ans << endl;
}

int main() {
    cin.tie(0);
   	ios::sync_with_stdio(false);
    solve();
    return 0;
}