Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define P pair<ll,ll>
#define FOR(I,A,B) for(ll I = ll(A); I < ll(B); ++I)
#define FORR(I,A,B) for(ll I = ll((B)-1); I >= ll(A); --I)
#define TO(x,t,f) ((x)?(t):(f))
#define SORT(x) (sort(x.begin(),x.end())) // 0 2 2 3 4 5 8 9
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin()) //xi>=v  x is sorted
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin()) //xi>v  x is sorted
#define NUM(x,v) (POSU(x,v)-POSL(x,v))  //x is sorted
#define REV(x) (reverse(x.begin(),x.end())) //reverse
ll gcd(ll a,ll b){if(a%b==0)return b;return gcd(b,a%b);}
ll lcm(ll a,ll b){ll c=gcd(a,b);return ((a/c)*(b/c)*c);}
#define NEXTP(x) next_permutation(x.begin(),x.end())
const ll INF=ll(1e18)+ll(7);
const ll MOD=1000000007LL;
#define out(a) cout<<fixed<<setprecision((a))


int main(){
	
	int n,m,k;
	cin >> n;
	vector<int> a(n);
	for(auto &x:a)cin >> x;
	cin >> m;
	vector<int> b(m);
	for(auto &x:b)cin >> x;
		b.push_back(MOD);
	for(auto x:a){
		int i = POSL(b,x);
		if(b[i] == x){
			cout << x << "\n";
		}
	}
}