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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <cassert>
#include <algorithm>
#include <functional>
#include <cmath>
#include <queue>
#include <set>
#include <stack>
#include <deque>
#include <map>
using ll = long long;
int MOD = 1e9 + 7;
using namespace std;
ll GCD(ll x, ll y)
{
    if (y == 0)
    {
        return x;
    }
    else
    {
        return GCD(y, x % y);
    }
}
ll LCM(ll x, ll y)
{
    return x / GCD(x, y) * y;
}
struct UnionFind
{
    vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

    UnionFind(int N) : par(N)
    { //最初は全てが根であるとして初期化
        for (int i = 0; i < N; i++)
            par[i] = i;
    }

    int root(int x)
    { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x)
            return x;
        return par[x] = root(par[x]);
    }

    void unite(int x, int y)
    {                     // xとyの木を併合
        int rx = root(x); //xの根をrx
        int ry = root(y); //yの根をry
        if (rx == ry)
            return;   //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
    }

    bool same(int x, int y)
    { // 2つのデータx, yが属する木が同じならtrueを返す
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
};
//使い方　インスタンス化(頂点数)　uniteつなぐ(双方向注意) same判定
int main()
{
    int n;
    cin >> n;
    vector<ll> vec(n);
    ll sum = 0;
    for (int i = 0; i < n;i++){
        cin >> vec[i];
        sum += vec[i];
    }
    bool h = false;
    ll c = 0;
    for (int i = 0; i < n;i++){
        if(h){
            if(vec[i]%2==0){
                c++;
                h = false;
            }else{
                h = false;
            }
        }else{
            if(vec[i]%2==1){
                h = true;
            }
        }
    }
    if(h){
        c++;
    }
    cout << (sum - c )/2<< endl;
}