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: Wrong Answer)

/*------------------------------------
........Bismillahir Rahmanir Rahim....
..........created by Abdul Aziz.......
------------------------------------*/
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <cmath>
#include <vector>
#include <set>
#include <list>
#include <stack>
#include <map>
#include <cstring>
#include <unordered_map>
#include <queue>
#define   mod  998244353
#define   int  long long 
#define   ld   long double
#define   pb   push_back
#define   vi  vector<int> 
#define   dbg(x)  cerr << #x << " = " << x << '\n'
#define   sz(x)  (int)x.size()
#define   all(a) (a.begin(),a.end())
#define   ff   first
#define   ss   second
#define   pii  pair<int,int> 
#define   lcm(a,b) (a*b)/__gcd(a,b) 
using namespace std;

inline void solve(){
    int n; cin>>n; 
    vi a(n+1);
    for (int i=1;i<=n;i++) cin>>a[i];
    int ans=0;
    for (int i=1;i<=n;i++){
        if (i!=n){
            if (a[i]>a[i+1]){
                int c = (a[i]-a[i+1]);
                ans += c/2;
                a[i] -= (c/2)*2;
            }
            int c=min(a[i],a[i+1]);
            ans += c;
            a[i+1] -= c;
        }
        else ans += a[i]/2;
    }
    cout << ans << endl;
}

signed main()
{
    int n=1; // cin>>n;
    while (n--) solve();
    return 0;
}