Codeforces 620D D. Professor Gukiz and Arrays

D. Professor Gukiz and Arrays

Professor Gukiz has both arrays of integers, A and B. Professor wants to make the sum of the elements in the array A s A as close as possible to the sum of the elements in the array b s C11>b. So he wants to minimize the value v = | Sa - sb|.

In one operation professor can swap some element from the Array a and some El Ement from the Array b. For example if the Array a is[5, 1, 3, 2, 4] and The Array b is [3, 3, 2] professor can Swap the Element 5 from the Array a and The Element 2 from the Array b and get The new Array a [2, 1, 3, 2, 4] and the new array  b [3, 3, 5].

Professor doesn ' t want to make more than and swaps. Find the minimal value V and some sequence of no more than-swaps that'll leads to the such value v. Professor makes swaps one by one, each new swap he makes with the new arrays A and B.

Input

The first line contains integer n (1≤ n ≤2000)-the number of elements in the array a.

The second line contains n integers ai ( -9≤ ai≤109)-the elements of the array a.

The third line contains integer m (1≤ m ≤2000)-the number of elements in the array b.

The fourth line contains m integers bJ ( -9≤ bJ≤109)-the elements of the array B.

Output

In the first line print the minimal value v = | Sa - sb| That can is got with no more than the swaps.

The second line should contain the number of swaps K (0≤ k ≤ 2).

Each of the next K lines should contain the integers xp, yp ( 1≤ xp ≤ n, 1≤ yp ≤ m)-the index of the element in the array A and the index of the element in the array b in the p-th swap.

If there is several optimal solutions print any of them. Print the swaps in order the professor did them.

Sample Test (s) input

5
5 4 3) 2 1
4
1 1 1 1

Output

1
2
1 1
4 2

Input

5
1 2 3) 4 5
1
15

Output

0
0

Input

5
1 2 3) 4 5
4
1 2 3 4

Output

1
1
3 1

#include <cstdio>#include<cstring>#include<map>#include<iostream>#include<algorithm>using namespaceStd;typedefLong LongLl;map<ll,pair<int,int> >F;inta[2][2005];ll sum[2];p Air<int,int>ans[2];intMain () {intn,m; scanf ("%d",&N);  for(intI=0; i<n;i++) scanf ("%d", &a[0][i]), sum[0]+=a[0][i]; scanf ("%d",&m);  for(intI=0; i<m;i++) scanf ("%d", &a[1][i]), sum[1]+=a[1][i]; if(sum[0]==sum[1]) {puts ("0\n0");return 0;} ll bound=abs (sum[0]-sum[1]);  for(intI=0; i<n;i++)         for(intj=0; j<m;j++) {ll TT=abs (sum[0]-2* (LL) a[0][i]+2* (LL) a[1][j]-sum[1]); if(Tt<bound) ans[0].first=i+1, ans[0].second=j+1, bound=tt; }     for(intI=0; i<n;i++)         for(intj=i+1; j<n;j++) f[2* (LL) a[0][i]+2* (LL) a[0][j]]=make_pair (i+1, j+1);  for(intI=0; i<m;i++)         for(intj=i+1; j<m;j++) {ll TT=sum[0]-sum[1]+2* (LL) a[1][i]+ (LL) (a[1][j])); Map<ll,pair<int,int> >::iterator iter=F.lower_bound (TT); if((Iter!=f.end () && (ABS (Tt-iter->first) <bound)) | | (Iter!=f.begin () && (tt-(--iter)->first) <bound)) {ans[0]=make_pair (iter->second.first,i+1); ans[1]=make_pair (iter->second.second,j+1); Bound=abs (iter->first-TT); }} printf ("%i64d\n", bound); if(bound==sum[0]-sum[1]) {puts ("0");return 0;} if(!ans[1].first) cout<<"1\n"<<ans[0].first<<" "<<ans[0].second<<Endl; Elsecout<<"2\n"<<ans[0].first<<" "<<ans[0].second<<endl<<ans[1].first<<" "<<ans[1].second<<Endl; return 0;}

Codeforces 620D D. Professor Gukiz and Arrays