#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define ld long double
#define pb push_back
#define pf push_front
#define pll pair<ll,ll>
#define pi 3.14159265358979323846
const ll MOD=1e9+7;
using namespace std;
ll POWER(ll base,ll expo)
{
    ll ret=1;
    while(expo)
    {
        if(expo&1)ret=ret*base;
        expo>>=1;
        base=base*base;
    }
    return ret;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ll test_case=1; //cin>>test_case;
    while(test_case--)
    {
        ll n1,n2,n3;cin>>n1>>n2>>n3;
        ll a1[n1],a2[n2],a3[n3];
        ll sum1=0,sum2=0,sum3=0;

        for(ll i=0;i<n1;i++){
            cin>>a1[i];
            sum1+=a1[i];
        }
        for(ll i=0;i<n2;i++){
            cin>>a2[i];
            sum2+=a2[i];
        }
        for(ll i=0;i<n3;i++){
            cin>>a3[i];
            sum3+=a3[i];
        }
        sort(a1,a1+n1);
        sort(a2,a2+n2);
        sort(a3,a3+n3);
        ll ans=-1e18;
        ans=max(ans,sum1+sum2-sum3);
        ans=max(ans,sum1+sum3-sum2);
        ans=max(ans,sum3+sum2-sum1);

        ans=max(ans,sum1+sum2+sum3-2*(a1[0]+a2[0]));
        ans=max(ans,sum1+sum2+sum3-2*(a1[0]+a3[0]));
        ans=max(ans,sum1+sum2+sum3-2*(a3[0]+a2[0]));

        cout<<ans;
    }
    return 0;
}

### Prompt

Please create a solution in cpp to the following problem:

You are given three bags. Each bag contains a non-empty multiset of numbers. You can perform a number of operations on these bags. In one operation, you can choose any two non-empty bags, and choose one number from each of the bags. Let's say that you choose number a from the first bag and number b from the second bag. Then, you remove b from the second bag and replace a with a-b in the first bag. Note that if there are multiple occurrences of these numbers, then you shall only remove/replace exactly one occurrence.

You have to perform these operations in such a way that you have exactly one number remaining in exactly one of the bags (the other two bags being empty). It can be shown that you can always apply these operations to receive such a configuration in the end. Among all these configurations, find the one which has the maximum number left in the end.

Input

The first line of the input contains three space-separated integers n_1, n_2 and n_3 (1 ≤ n_1, n_2, n_3 ≤ 3⋅10^5, 1 ≤ n_1+n_2+n_3 ≤ 3⋅10^5) — the number of numbers in the three bags.

The i-th of the next three lines contain n_i space-separated integers a_{{i,1}}, a_{{i,2}}, ..., a_{{i,{{n_i}}}} (1 ≤ a_{{i,j}} ≤ 10^9) — the numbers in the i-th bag.

Output

Print a single integer — the maximum number which you can achieve in the end.

Examples

Input


2 4 1
1 2
6 3 4 5
5


Output


20

Input


3 2 2
7 5 4
2 9
7 1


Output


29

Note

In the first example input, let us perform the following operations:

[1, 2], [6, 3, 4, 5], [5]

[-5, 2], [3, 4, 5], [5] (Applying an operation to (1, 6))

[-10, 2], [3, 4], [5] (Applying an operation to (-5, 5))

[2], [3, 4], [15] (Applying an operation to (5, -10))

[-1], [4], [15] (Applying an operation to (2, 3))

[-5], [], [15] (Applying an operation to (-1, 4))

[], [], [20] (Applying an operation to (15, -5))

You can verify that you cannot achieve a bigger number. Hence, the answer is 20.

### Response

```cpp
#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define ld long double
#define pb push_back
#define pf push_front
#define pll pair<ll,ll>
#define pi 3.14159265358979323846
const ll MOD=1e9+7;
using namespace std;
ll POWER(ll base,ll expo)
{
    ll ret=1;
    while(expo)
    {
        if(expo&1)ret=ret*base;
        expo>>=1;
        base=base*base;
    }
    return ret;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ll test_case=1; //cin>>test_case;
    while(test_case--)
    {
        ll n1,n2,n3;cin>>n1>>n2>>n3;
        ll a1[n1],a2[n2],a3[n3];
        ll sum1=0,sum2=0,sum3=0;

        for(ll i=0;i<n1;i++){
            cin>>a1[i];
            sum1+=a1[i];
        }
        for(ll i=0;i<n2;i++){
            cin>>a2[i];
            sum2+=a2[i];
        }
        for(ll i=0;i<n3;i++){
            cin>>a3[i];
            sum3+=a3[i];
        }
        sort(a1,a1+n1);
        sort(a2,a2+n2);
        sort(a3,a3+n3);
        ll ans=-1e18;
        ans=max(ans,sum1+sum2-sum3);
        ans=max(ans,sum1+sum3-sum2);
        ans=max(ans,sum3+sum2-sum1);

        ans=max(ans,sum1+sum2+sum3-2*(a1[0]+a2[0]));
        ans=max(ans,sum1+sum2+sum3-2*(a1[0]+a3[0]));
        ans=max(ans,sum1+sum2+sum3-2*(a3[0]+a2[0]));

        cout<<ans;
    }
    return 0;
}

```