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

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

Output

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

Example

Input

5
1 2 3 5 8
2
2 5


Output

1
3
8

Solution (in CPP):

Solution:
#include<iostream>
#include<set>
using namespace std;

int main()
{
    set<int> s1, s2;
    int n, m;

    cin >> n;
   
    int a;

    for( int i = 0; i < n; i++ )
    {
        cin >> a;
        s1.insert(a);
    }

    cin >> m;

    for( int i = 0; i < m; i++ )
    {    
        cin >> a;
        s2.insert(a);
    }

    set<int>::iterator itr1 = s1.begin(), itr2 = s2.begin();
    for(; itr1 != s1.end() && itr2 != s2.end(); )
    {
        if( *itr1 < *itr2 )
        {
            cout << *itr1 << endl;
            itr1++;
        }
        else if( *itr1 > *itr2 )        //s2の集合の値は小さくても出力しない。s1の要素からs2の要素を省いたものが差集合だから。
            itr2++;
        else
        {    
            itr1++;
            itr2++;
        }
    }

    for(; itr1 != s1.end(); itr1++ )    //前のfor分を抜けた時、もとの集合の要素が余る場合（s2の集合の最大要素よりも大きい要素がある時）、全部出力する必要がある。
        cout << *itr1 << endl;

    return 0;
}