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 <iostream>
#include <iomanip>
#include <string>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <list>
#include <stack>
#include <queue>
#include <bitset>
#include <numeric>
#include <cassert>
#ifdef DEBUG
#include "./Lib/debug.hpp"
#else
#define dump(...)
template<class T>inline auto d_val(T a, T b) { return a; }
#endif

/* (=＾o＾=) */
#define int ll

/* macro */
#define FOR(i, b, e) for(ll i = (ll)(b); i < (ll)(e); ++i)
#define RFOR(i, b, e) for(ll i = (ll)(e-1); i >= (ll)(b); --i)
#define REP(i, n) FOR(i, 0, n)
#define RREP(i, n) RFOR(i, 0, n)
#define REPC(x,c) for(const auto& x:(c))
#define REPI2(it,b,e) for(auto it = (b); it != (e); ++it)
#define REPI(it,c) REPI2(it, (c).begin(), (c).end())
#define RREPI(it,c) REPI2(it, (c).rbegin(), (c).rend())
#define REPI_ERACE2(it, b, e) for(auto it = (b); it != (e);)
#define REPI_ERACE(it, c) REPI_ERACE2(it, (c).begin(), (c).end())
#define ALL(x) (x).begin(),(x).end()
#define cauto const auto&
/* macro func */
#define SORT(x) do{sort(ALL(x));}while(false)
#define RSORT(x) do{sort((x).rbegin(),(x).rend());}while(false)
#define UNIQUE(v) do{v.erase( unique(v.begin(), v.end()), v.end() );}while(false)
#define MAX(x,y) do{x = std::max(x,y);}while(false)
#define MIN(x,y) do{x = std::min(x,y);}while(false)
#define BR do{cout<<"\n";}while(false)

/* type define */
using ll = long long;
using PAIR = std::pair<ll, ll>;
using VS = std::vector<std::string>;
using VL = std::vector<long long>;
using VVL = std::vector<VL>;
using VVVL = std::vector<VVL>;
using VD = std::vector<double>;
template<class T>
using V = std::vector<T>;

/* using std */
using std::cout;
constexpr char endl = '\n';
using std::cin;
using std::sort;
using std::pair;
using std::string;
using std::stack;
using std::queue;
using std::vector;
using std::list;
using std::map;
using std::unordered_map;
using std::multimap;
using std::unordered_multimap;
using std::set;
using std::unordered_set;
using std::unordered_multiset;
using std::multiset;
using std::bitset;
using std::priority_queue;

/* constant value */
constexpr ll MOD = 1000000007;
//constexpr ll MOD = 998244353;

/* Initial processing  */
struct Preprocessing { Preprocessing() { std::cin.tie(0); std::ios::sync_with_stdio(0); }; }_Preprocessing;

/* Remove the source of the bug */
signed pow(signed, signed) { assert(false); return -1; }

/* define hash */
namespace std { template <>	class hash<std::pair<ll, ll>> { public:	size_t operator()(const std::pair<ll, ll>& x) const { return hash<ll>()(1000000000 * x.first + x.second); } }; }

/* input */
template<class T> std::istream& operator >> (std::istream& is, vector<T>& vec) { for (T& x : vec) is >> x; return is; }

//=============================================================================================


signed main() {
	ll n;
	cin >> n;
	VL v(n);
	cin >> v;

	if (v[0] > 0) { cout << -1 << endl; return 0; }

	ll ans = 0;
	REP(i, n - 1) {
		if (v[i + 1] > v[i] + 1) { cout << -1 << endl; return 0; }
		if (v[i + 1] == v[i] + 1) { ++ans; }
		if (v[i + 1] <= v[i]) {
			ans += v[i + 1];
		}
	}
	cout << ans << endl;
}