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 ll;
typedef long double ld;
using namespace std;

using Pii = pair<int, int>;
using Pll = pair<ll, ll>;

#define REP(i, l, n) for(int i=(l), i##_len=(n); i<i##_len; ++i)
#define ALL(x) (x).begin(),(x).end()
#define pb push_back

ll gcd(ll a,ll b){return b ? gcd(b,a%b) : a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}

char alpha[26] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'};

int dx[4] = {-1, 1, 0, 0};
int dy[4] = {0, 0, 1, -1};

int main(){
    int n; cin >> n;
    vector<ll> alis(n);
    REP(i, 0, n){cin >> alis[i];}
    ll key = -1;
    REP(i, 0, n){
        if(key+1 == alis[i]){
            key++;
        }else if(key >= alis[i]){
            key = alis[i];
        }else{
            cout << -1 << endl;
            return 0;
        }
    }
    ll ans = alis[n-1];
    for(int i = n-2; i >= 0; i--){
        if(alis[i]+1 == alis[i+1]){
            continue;
        }else{
            ans += alis[i];
        }
    }
    cout << ans << endl;
}