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 <stdio.h>
#include <algorithm>
#include <iostream>
#include <limits.h>
#include <math.h>
#include <queue>
#include <set>
#include <stdlib.h>
#include <string>
#include <vector>
#include <cstdio>
#include <iomanip>
#define ll long long
#define rep2(i,a,b) for(int i=a;i<=b;i++)
#define rep(i,n) for(int i=0;i<n;i++)
#define rep3(i,a,b) for(int i=a;i>=b;i--)
#define REP(e,v) for(auto e:v)
#define queint queue<int>
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pq priority_queue<int>
#define pqg priority_queue<int,vector<int>,greater<int>()>
#define pb push_back
#define vec vector<int>
#define vecvec vector<vector<int>>
#define vecll vector<ll>
#define vecvecll vector<vector<ll>>
#define bs binary_search
#define All(c) (c).begin(),(c).end()
#define mp make_pair
using namespace std;
int in(){int x;scanf("%d",&x);return x;}
string stin(){string s;cin>>s;return s;}
ll lin(){ll x;scanf("%lld",&x);return x;}
int main(){
    int n=in();
    vec a(n);
    rep(i,n)a[i]=in();
    if(a[0]!=0){
        cout<<-1<<endl;
        return 0;
    }
    int id=n-1,tmp=a[n-1];
    ll cnt=a[n-1];
    while(id>0){
        if(a[--id]==tmp-1){
            tmp--;
            continue;
        }else{
            if(tmp>a[id]+1){
                cout<<-1<<endl;
                return 0;
            }
            cnt+=a[id];
            tmp=a[id];
        }
    }
    cout<<cnt<<endl;
}