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)

//code by lynmisakura.wish to be accepted!
/****************************/
#include<iostream>
#include<iomanip>
#include<math.h>
#include<vector>
#include<string>
#include<stack>
#include<queue>
#include<map>
#include<algorithm>
#include<bitset>
#include <climits>
#include<set>
#include<bitset>
using namespace std;
/***************************/
typedef long long ll;
typedef vector<ll> vi;
typedef vector<long long> vl;
typedef pair<ll, ll> pi;
typedef vector<pair<ll, ll>> vpi;

//const long long INF = 1LL << 55;

#define itn ll
#define endl '\n'
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define ss second
#define ff first
#define dup(x,y) ((x) + (y) - 1)/(y)
#define mins(x,y) x = min(x,y)
#define maxs(x,y) x = max(x,y)
#define all(x) (x).begin(),(x).end()
#define Rep(n) for(ll i = 0;i < n;i++)
#define rep(i,n) for(ll i = 0;i < n;i++)
#define flush fflush(stdout)
#define rrep(i,n) for(ll i = n - 1;i >= 0;i--)
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() )
#define ioboost cin.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(20)
ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
ll qp(ll a, ll b) { ll ans = 1; do { if (b & 1)ans = 1ll * ans*a; a = 1ll * a*a; } while (b >>= 1); return ans; }
ll qp(ll a, ll b, ll mo) { ll ans = 1; do { if (b & 1)ans = 1ll * ans*a%mo; a = 1ll * a*a%mo; } while (b >>= 1); return ans; }

#define _GLIBCXX_DEBUG
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl

int main(void){
    int n;cin >> n;
    vector<ll> a(n);rep(i,n)cin >> a[i];
    reverse(all(a));
    long long ans = a[0];
    for(int i = 0;i < n;i++){
        if(a[i] > n - 1 - i){
            cout << -1 << endl;
            return 0;
        }else if(i){
            if(a[i-1] - a[i] > 1){
                cout << -1 << endl;
                return 0;
            }else if(a[i-1] - a[i] == 1){
                continue;
            }else{
                ans += a[i];
            }
        }
    }
    cout << ans << endl;
}