Problem:
Score : 400 points

Problem Statement

Snuke has a large collection of cards. Each card has an integer between 1 and N, inclusive, written on it.
He has A_i cards with an integer i.

Two cards can form a pair if the absolute value of the difference of the integers written on them is at most 1.

Snuke wants to create the maximum number of pairs from his cards, on the condition that no card should be used in multiple pairs. Find the maximum number of pairs that he can create.

Constraints

1 ≦ N ≦ 10^5

0 ≦ A_i ≦ 10^9 (1 ≦ i ≦ N)

All input values are integers.

Input

The input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

Print the maximum number of pairs that Snuke can create.

Sample Input 1

4
4
0
3
2

Sample Output 1

4

For example, Snuke can create the following four pairs: (1,1),(1,1),(3,4),(3,4).

Sample Input 2

8
2
0
1
6
0
8
2
1

Sample Output 2

9

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>

using namespace std;

using i64 = long long;
#define forn(a, e) for (i64 a = 0; a < (i64)(e); a++)
#define forr(a, s, e) for (i64 a = s; a < (i64)(e); a++)
#define fore(e, a) for (auto& e : a)

#ifdef LOCAL
#define logv(a) {cerr << #a << " = "; fore(e, a) {cerr << e << " ";} cerr << "\n";}
#define logvp(a) {cerr << #a << " = "; fore(e, a) {cerr << "(" << e.first << ", " << e.second << ") ";} cerr << "\n";}
#define logvv(a) {cerr << #a << " = \n"; fore(r, a) { fore(e, r) {cerr << e << " ";} cerr << "\n";} }
#define logvf(a, field) {cerr << #a"."#field << " = \n"; fore(e, a) { cerr << e.field << " ";} cerr << "\n"; }
#define logvff(a, f1, f2) {cerr << #a".{"#f1 << ", "#f2 << "} = \n"; fore(e, a) { cerr << "(" << e.f1 <<", " << e.f2 << ") ";} cerr << "\n"; }
#define logs(a) cerr << #a << " = " << (a) << "\n";
#define logss(a, b) cerr << #a << " = " << (a) << ", " << #b << " = " << (b) << "\n";
#define logp(a) cerr << #a << " = " << "(" << a.first << ", " << a.second << ")" << "\n";
#define cond(pred, stmt) if (pred) { stmt }
#else
#define logv(a)
#define logvp(a)
#define logvv(a)
#define logvf(a, field)
#define logvff(a, f1, f2)
#define logs(a)
#define logss(a, b)
#define logp(a)
#define cond(pred, stmt)
#endif

using iip = pair<int, int>;
using llp = pair<i64, i64>;
using ivec = vector<int>;
using llvec = vector<i64>;
using svec = vector<string>;
template<typename T> using vec = vector<T>;

template<typename T, typename Dim>
auto make_vec(T value, Dim dim) { return vector<T>(dim, value); }
template<typename T, typename Dim1, typename... Dim>
auto make_vec(T value, Dim1 dim1, Dim... dims) { return make_vec(make_vec(value, dims...), dim1); }

template<typename T>
bool uax(T& v, const T& newv) { if (v < newv) { v = newv; return true; } else return false; }
template<typename T>
bool uin(T& v, const T& newv) { if (v > newv) { v = newv; return true; } else return false; }

template<typename T>
istream& operator>>(istream& is, vector<T>& c) { for (auto& e : c) is >> e; return is; }
template<typename T, size_t N>
istream& operator>>(istream& is, array<T, N>& c) { for (auto& e : c) is >> e; return is; }

template<typename ...T>
istream& read(T&... args) { return (cin >> ... >> args); }

static mt19937 rande(123123);
template<typename T>
T rand_int(T from, T to) { uniform_int_distribution<T> distr(from, to); return distr(rande); }

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    rande.seed(chrono::steady_clock::now().time_since_epoch().count());

    int n;
    while (read(n)) {
        llvec a(n);
        read(a);

        auto dp = make_vec(-1ll, n, 2);
        dp[0][a[0] % 2] = a[0] / 2;
        forr(i, 1, n) {
            if (a[i] % 2 == 0) {
                dp[i][0] = max(dp[i - 1][0], dp[i - 1][1]) + a[i] / 2;
                dp[i][1] = -1;
                if (dp[i - 1][1] != -1 && a[i] > 0) {
                    dp[i][1] = dp[i - 1][1] + a[i] / 2;
                }
            } else {
                dp[i][1] = max(dp[i - 1][0], dp[i - 1][1]) + a[i] / 2;
                dp[i][0] = -1;
                if (dp[i - 1][1] != -1) {
                    dp[i][0] = dp[i - 1][1] + a[i] / 2 + 1;
                }
            }
        }
        logv(a);
        logvv(dp);
        cout << max(dp[n - 1][0], dp[n - 1][1]) << endl;
    }
}