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)

#include<bits/stdc++.h>
using namespace std;

#define int long long 
#define ld long double
#define pii pair<int, int>
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define fi first
#define se second
#define dbg(x) cerr << #x << ": " << x << '\n'
#define fast_io() ios::sync_with_stdio(false); cin.tie(0);

signed main() {
    fast_io();
    int n;
    cin >> n;
    vector<int> a(n);
    vector<vector<int>> here(n);
    for(int i = 0; i < n; i++) cin >> a[i];
    for(int i = 0; i < n; i++){
        if(a[i] > (i == 0 ? 0 : a[i - 1] + 1)){
            cout << -1;
            return 0;
        }
        here[i - a[i]].pb(a[i] - i);
    }
    map<int, int> mp;
    int cnt = 0, ans = 0;
    for(int i = 0; i < n; i++){
        for(auto x : here[i]){
            mp[x]++;
            if(mp[x] == 1) cnt++;
        }
        ans += cnt - min(1LL, mp[-i]);
        mp[a[i] - i]--;
        if(mp[a[i] - i] == 0) cnt--;
    }
    cout << ans;
    return 0;
}