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>
#define int long long
#define P(x) {if (debug) cout << x << endl;}
#define H(x) P(#x << ": " << (x))
#define FR(i,a,b) for (int i=(a); i<(b); i++)
#define F(i,n) FR(i,0,n)
#define DR(i,a,b) for (int i=(b); i-->(a);)
#define D(i,n) DR(i,0,n)
#define S(s) (int)(s).size()
#define ALL(v) v.begin(), v.end()
#define MI(a,v) a = min(a,v)
#define MA(a,v) a = max(a,v)
#define V vector
#define pb push_back
#define mt make_tuple
using namespace std;
template<class T> ostream &operator<<(ostream &os, V<T> v) {
    F(i,S(v)) os<<(i?" ":"")<<v[i];
    return os;
}
const bool debug = 1;

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int n; cin>>n;
    V<int> a(n);
    F(i,n) cin>>a[i];
    int cost = 0;
    bool poss = (a[0] == 0);
    D(i,n) {
        if (i+1<n && a[i+1] > a[i]+1)
            poss = false;
        if (i+1 == n || a[i] > a[i+1]-1)
            cost += a[i];
    }
    if (!poss) cost = -1;
    cout<<cost<<"\n";
}