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++ (GCC 9.2.1) (judge verdict: Accepted)

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>; 
using pll = pair<long long, long long>;
constexpr char ln =  '\n';
constexpr long long MOD = 1000000007LL;
constexpr long long INF = 1001001001LL;
constexpr long long LINF = 1001001001001001001;
#define all(x) (x).begin(),(x).end()
#define rep(i,n) for(int i=0;i<(n);i++)
#define rept(i, j, n) for(int i=(j); i<(n); i++)
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }


int main(){
    int n; cin >> n;
    vector<ll> A(n); rep(i, n) cin >> A[i];
    
    bool fg = true;
    if(A[0] != 0)fg = false;
    int bef = A[0]; ll sum = 0;
    rept(i, 1, n){
      if(A[i] > bef + 1){fg = false; break;}
      if(A[i] < bef + 1){
        sum += A[i-1];
      }
      bef = A[i];
    }
    sum += A[n-1];
    if(fg) cout << sum << ln;
    else cout << -1 << ln;
}