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;
typedef long long ll;
typedef pair<int,int> P;
typedef pair<int,P> P1;
typedef pair<P,P> P2;
#define pu push
#define pb push_back
#define mp make_pair
#define eps 1e-7
#define INF 1000000000
#define mod 1000000007
#define fi first
#define sc second
#define rep(i,x) for(int i=0;i<x;i++)
#define repn(i,x) for(int i=1;i<=x;i++)
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())
int n,a[200005];
int main(){
	scanf("%d",&n);
	repn(i,n) scanf("%d",&a[i]);
	if(a[1] != 0){
		puts("-1"); return 0;
	}
	repn(i,n-1){
		if(a[i+1]-a[i] >= 2){
			puts("-1"); return 0;
		}
	}
	ll ans = 0;
	set<int>zip;
	int lazy = 0;
	for(int i=n;i>=2;i--){
		if(zip.find(-lazy) != zip.end()){
			zip.erase(zip.find(-lazy));
		}
		if(zip.find(1-lazy) != zip.end()){
			zip.erase(zip.find(1-lazy));
		}
		if(zip.find(a[i]+1-lazy) != zip.end()){
			zip.erase(zip.find(a[i]+1-lazy));
		}
		lazy--;
		if(a[i] != 0){
		    zip.insert(a[i]-lazy);
	    	ans += zip.size();
		}
	}
	cout<<ans<<endl;
}