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: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define NDEBUG
#ifdef DEBUG
#include "../cout11.h"
#undef NDEBUG
#endif
#include <cassert>

typedef long long ll;
typedef long double Double;
typedef unsigned long long ull;
typedef pair<int,int> ii;
typedef pair<ll,ll> llll;
typedef pair<double,double> dd;

typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<ii> vii;
typedef vector<vector<ii>> vvii;
typedef vector<ll> vll;
typedef vector<string> vs;
typedef vector<double> vd;
typedef vector<long double> vD;

#define sz(a)  int((a).size())
#define pb  push_back
#define FOR(var,from,to) for(int var=(from);var<=(to);++var)
#define rep(var,n)  for(int var=0;var<(n);++var)
#define rep1(var,n)  for(int var=1;var<=(n);++var)
#define repC2(vari,varj,n)  for(int vari=0;vari<(n)-1;++vari)for(int varj=vari+1;varj<(n);++varj)
#define ALL(c)  (c).begin(),(c).end()
#define RALL(c)  (c).rbegin(),(c).rend()
#define tr(i,c)  for(auto i=(c).begin(); i!=(c).end(); ++i)
#define found(s,e)  ((s).find(e)!=(s).end())
#define mset(arr,val)  memset(arr,val,sizeof(arr))
#define mid(x,y) ((x)+((y)-(x))/2)
#define IN(x,a,b) ((a)<=(x)&&(x)<=(b))
#define cons make_pair


ll gcd(ll a, ll b) { while(a) swap(a, b%=a); return b; }

const ll MOD=1000000007LL;

ll ADD(ll x, ll y) { return (x+y) % MOD; }
ll SUB(ll x, ll y) { return (x-y+MOD) % MOD; }
ll MUL(ll x, ll y) { return x*y % MOD; }
ll POW(ll x, ll e) { ll v=1; for(; e; x=MUL(x,x), e>>=1) if (e&1) v = MUL(v,x); return v; }
ll DIV(ll x, ll y) { /*assert(y%MOD!=0);*/ return MUL(x, POW(y, MOD-2)); }
// ll comb(ll n, ll k) { ll v=1; for(ll i=1; i<=k; i++) v = DIV(MUL(v, n-i+1),i); return v; }
//


ll solve(int N, vi& a) {
    ll rem0=-1, rem1=-1;

    rem0 = a[0]/2;
    if (a[0]%2) rem1 = rem0;

    for (int i=1; i<N; ++i) {
#ifdef DEBUG
        // fprintf(stderr, "%d] %lld %lld\n", i-1, rem0, rem1);
#endif
        ll nrem0=-1, nrem1=-1, tmp;
        if (rem0 >= 0) {
            tmp = rem0 + a[i]/2;
            if (a[i] % 2) nrem1 = max(nrem1, tmp);
            else          nrem0 = max(nrem0, tmp);
        }
        if (rem1 >= 0) {
            if (a[i]) {
                tmp = rem1 + 1 + (a[i]-1)/2;
                if ((a[i]-1) % 2) nrem1 = max(nrem1, tmp);
                else              nrem0 = max(nrem0, tmp);
            }
            tmp = rem1 + a[i]/2;
            if (a[i] % 2) nrem1 = max(nrem1, tmp);
            else          nrem0 = max(nrem0, tmp);
        }
        rem0 = nrem0; rem1 = nrem1;
    }
#ifdef DEBUG
    // fprintf(stderr, "%d] %lld %lld\n", N-1, rem0, rem1);
#endif
    return max(rem0, rem1);
}

int main() {
    int N; cin>>N;
    vi a(N); rep(i,N) cin>>a[i];
    cout << solve(N,a) << endl;
    return 0;
}