// Problem : B - An Odd Problem
// Contest : AtCoder - AIsing Programming Contest 2020
// URL : https://atcoder.jp/contests/aising2020/tasks/aising2020_b
// Memory Limit : 1024 MB
// Time Limit : 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cp-editor)

#include <bits/stdc++.h>
using namespace std;
 
using ll = long long;
using ull = unsigned long long;
using db = double;
using pii = pair < int, int >;
using pll = pair < ll, ll >;
using mapl = map < ll , ll >;
using mapc = map < char , ll >;
using vectorl = vector < ll >;
using vectors = vector < string >;
const int N = 2e5 + 10;
const ll inf = 1e15 + 42;
#define pb push_back
#define endl "\n"
const ll mod = 1000000007;

ll i=0,j=0,k=0,t=0,n=0;

char x='a',y='a',z='a';
string s="",s1="";

vectorl pr;
void seive(int n){ 
    bool prime[n+1]; 
    memset(prime, true, sizeof(prime)); 
  
    for (int p=2; p*p<=n; p++){ 
        if (prime[p] == true){ 
            for (int i=p*p; i<=n; i += p) 
                prime[i] = false; 
        } 
    } 
    for (int p=2; p<=n; p++) 
      if (prime[p]) 
          pr.push_back(p);
} 

double log_base(ll x, ll base) {
    return ((log(x)*1.0) / (log(base)*1.0));
}

ll calcInverse(ll a, ll n)
{
    ll t = 0, newt = 1;
    ll r = n, newr = a;  
    while (newr != 0) {
        auto quotient = r /newr;
        tie(t, newt) = make_tuple(newt, t- quotient * newt);
        tie(r, newr) = make_tuple(newr, r - quotient * newr);
    }
    if (r > 1)
        throw runtime_error("a is not invertible");
    if (t < 0)
        t += n;
        
    t=t%mod;    
    return t;
}

ll power_without_mod(ll a,ll b){
    ll ans=1;
    while(b>0){
        if(b&1){
            ans=ans*a;
        }
        a = a * a;
        b >>= 1;
    }
    return ans;
}

ll power(ll a,ll b){
	a%=mod;
	ll ans=1;
	while(b>0){
		if(b&1){
			ans=ans*a%mod;
		}
		a = a * a%mod;
		b >>= 1;
	}
	return ans;
}

// if mod is prime: calculate x^(n mod (m-1))
// if  is string   find number calculating with mod of (m-1)

//for ex:
// string b = "100000000000000000000000000"; 
  
//     ll remainderB = 0; 
//     ll MOD = 1000000007; 
  
//     // Reduce the number B to a small number 
//     // using Fermat Little 
//     for (int i = 0; i < b.length(); i++) 
//         remainderB = (remainderB * 10 + b[i] - '0') % (MOD - 1); 
  
//     cout << power(a, remainderB, MOD) << endl; 


void multiply(ll a[3][3], ll b[3][3]) 
{ 
    // Creating an auxiliary matrix to store elements  
    // of the multiplication matrix 
    ll mul[3][3]; 
    for (ll i = 0; i < 3; i++) 
    { 
        for (ll j = 0; j < 3; j++) 
        { 
            mul[i][j] = 0; 
            for (ll k = 0; k < 3; k++) 
                mul[i][j] = (mul[i][j]+((a[i][k]%mod)*(b[k][j]%mod))%mod)%mod; 
        } 
    } 
  
    // storing the multiplication result in a[][] 
    for (ll i=0; i<3; i++) 
        for (ll j=0; j<3; j++) 
            a[i][j] = mul[i][j];  // Updating our matrix 
} 
  
// Function to compute F raise to power n-2. 
ll power_matrix(ll F[3][3],ll n) 
{ 
    ll result[3][3] = {{1,0,0}, {0,1,0}, {0,0,1}};
    while(n>0){
    	if(n%2==1){
    		multiply(result,F);
    	}
    	multiply(F,F);
    	n/=2;
    }
    return result[0][0]*3+result[0][2];
} 



void solve(){
	ll n;
	cin>>n;
	ll a[n];
	for(i=0;i<n;i++)
		cin>>a[i];
	ll co=0;
	for(i=0;i<n;i+=2){
		if(a[i]%2!=0)
			co++;
	}
	cout<<co<<endl;
}

int main() {
    ios_base :: sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    // ll t;
    // cin>>t;
    // while(t--){
    	solve();
    // }
    return 0;
}





Measured memory_bytes: 3604