Find the intersection of two sets $A = \\{a_0, a_1, ..., a_{n-1}\\}$ and $B = \\{b_0, b_1, ..., b_{m-1}\\}$.

Constraints

* $1 \leq n, m \leq 200,000$
* $0 \leq a_0 < a_1 < ... < a_{n-1} \leq 10^9$
* $0 \leq b_0 < b_1 < ... < b_{m-1} \leq 10^9$

Input

The input is given in the following format.


$n$
$a_0 \; a_1 \; ... \; a_{n-1}$
$m$
$b_0 \; b_1 \; ... \; b_{m-1}$


Elements of $A$ and $B$ are given in ascending order respectively. There are no duplicate elements in each set.

Output

Print elements in the intersection in ascending order. Print an element in a line.

Example

Input

4
1 2 5 8
5
2 3 5 9 11


Output

2
5
#include <iostream>
#include <string>
#include <vector>
#include <algorithm> 
#include <utility> 
#include <tuple> 
#include <cstdint> 
#include <cstdio> 
#include <map> 
#include <queue> 
#include <set> 
#include <stack> 
#include <deque> 
#include <unordered_map> 
#include <unordered_set> 
#include <bitset> 
#include <cctype> 
#include <cmath>
#include <iomanip>
#include <ctype.h>
#include <numeric>
#include <iterator>

using namespace std;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi >;
using vl = vector<ll>;
using vvl = vector<vl >;
using pairi = pair<int, int>;
using pairl = pair<ll, ll>;

#define TR ","
#define TS " "
#define rep(i,N) for(ll i=0;i<(ll)N;++i)
#define rpt(i,a,b) for(ll i=a;i<(ll)b;++i)
#define all(v) v.begin(), v.end()
#define IO ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)

ll gcd(ll a, ll b) {
    if (a < b) swap(a, b);
    if (b == 0) return a;
    return gcd(b, a % b);
}

ll lcm(ll x, ll y) {
    return x / gcd(x, y) * y;
}

ll waz = 76543217;

void printVector(const vector<int>& vec) {
    for (int value : vec) {
        cout << value << " ";
    }
    cout << endl;
}

int main()
{
    IO;

    ll k, n, m;
    set<ll> a, b;
    set<ll> result;

    cin >> n;
    rep(i, n) {
        cin >> k;
        a.insert(k);
    }

    cin >> m;
    rep(i, m) {
        cin >> k;
        b.insert(k);
    }

    set_intersection(a.begin(), a.end(),
        b.begin(), b.end(),
        inserter(result, result.end()));

    for_each(result.begin(), result.end(), [](ll x) {
        cout << x << endl;
        });

}
