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>
using namespace std;
#define ll          long long int
#define pb          push_back
#define pll         pair<ll,ll>
#define speed_up    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define mode        998244353
#define mod         1000000007
#define inf         1000000000000000000LL
#define vl          vector<ll>
#define F           first
#define S           second
#define db          long double
#define sz(x)       (ll)x.size()
#define fix(n)      cout<<fixed<<setprecision(n)
#define rep(i,a,b)  for(ll i=a;i<b;i++)
#define all(x)      x.begin(),x.end()
#define mset(x)     memset(x,0,sizeof x)
#define pi          3.14159265358979323

const int N = 2e5+5;
ll n, a[N], dp[N], ans=0;

void solve()
{
       ll n; cin>>n;
       rep(i,1,n+1) cin>>a[i];
       rep(i,1,n+1)
       {
              //cin>>a[i];
              if( (i==1 && a[i]) || (a[i]-a[i-1]>1)) { cout<<-1; exit(0); }
              if(a[i]==a[i-1]+1) ans++;
              else ans+=a[i];
       }
       cout<<ans;
}


int main()
{
    speed_up
    ll T=1;
    //cin>>T;
    while(T--)
        solve();
    return 0;
}