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;
using ll = long long;
using P = pair<int, int>;
#define PI acos(-1)
#define Mod (int)1000000007
#define INFTY (int)INT_MAX
#define Rep(i, n) for (int i = 0; i < (int)(n); i++)
#define BitRep(i, n) for (int i = 0; i < (int)(1 << n); i++)
#define All(vec) vec.begin(), vec.end()
#define Part(vec, b, e) vec.begin() + b, vec.begin() + e
#define Sum(vec) accumulate(vec.begin(), vec.end(), 0)
#define Sort(vec) sort(vec.begin(), vec.end())
#define Reverse(vec) reverse(vec.begin(), vec.end())
#define Count(vec, x) count(vec.begin(), vec.end(), x)
#define Next_permutation(vec) next_permutation(vec.begin(), vec.end())
#define Find(vec, s) vec.find(s) != vec.end()

int main() {
	set<int> A, B;
	int n;
	cin >> n;
	Rep(i, n) {
		int a;
		cin >> a;
		A.insert(a);
	}
	int m;
	cin >> m;
	Rep(i, m) {
		int b;
		cin >> b;
		B.insert(b);
	}
	set<int> res;
	set_intersection(All(A), All(B), inserter(res, res.end()));
	for (auto v : res) {
		cout << v << endl;
	}
	return 0;
}