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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
struct node{
    int s,e,m,v,la;
    node  *l, *r;
    node (int _s, int _e): s(_s), e(_e){
        m = (s+e)/2;
        la = 0;
        if (s == e){
            v = -1;
        }
        else{
            l = new node(s,m);
            r = new node(m+1,e);
            v = l->v + r->v;
        }
    }
    void prop(){
        if (la){
            v = 0;
            if (s != e ){
                l -> la = 1;
                r -> la = 1;
            }
            la = 0;
        }
    }
    void up(int qs, int qe){
        prop();
        if (qs == s && qe == e){
            la = 1;
            return;
        }
        if (qs  > m) r->up(qs,qe);
        else if (qe <= m)  l->up(qs,qe);
        else l->up(qs,m), r->up(m+1,qe);
        l->prop(); r->prop();
        v = l->v+r->v;
    }
    int qu(int qs, int qe){
        prop();
        if (qs == s && qe == e)  return v;
        if (qs > m) return r->qu(qs,qe);
        else if (qe <= m) return l->qu(qs,qe);
        else return l->qu(qs,m) + r->qu(m+1,qe);
    }
} *root;
int n;
int a[200005];

int main(){
    scanf("%d",&n);
    for (int i = 1; i <= n; i++){
        scanf("%d",&a[i]);
        if (a[i] >= i){
            printf("-1\n");
            return 0;
        }
    }
    root = new node(1,n);
    ll ans = 0;
    for (int i = n; i >= 1; i--){
        int expect = i+root->qu(1,i);
        if (expect != a[i]){
            ans += a[i];
            root->up(i-a[i]+1,i);
        }
    }
    printf("%lld",ans);
}