#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<int> vint;
typedef pair<int,int> pint;
typedef vector<pint> vpint;
#define rep(i,n) for(int i=0;i<(n);i++)
#define reps(i,f,n) for(int i=(f);i<(n);i++)
#define each(it,v) for(__typeof((v).begin()) it=(v).begin();it!=(v).end();it++)
#define all(v) (v).begin(),(v).end()
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define chmax(a, b) a = (((a)<(b)) ? (b) : (a))
#define chmin(a, b) a = (((a)>(b)) ? (b) : (a))
const int MOD = 1e9 + 7;
const int INF = 1e9;

int n;
int L[30];
int t;
string name[4010];
char P[4010];
int cnt[30];

struct hito
{
	string na;
	int sum = 0;
	int kousin;
	int score[30] = {0};
};
bool asc( const hito& left, const hito& right ) {
    return left.sum == right.sum ? left.kousin < right.kousin : left.sum > right.sum;
}

int main(void){
	cin >> n;
	rep(i, n) cin >> L[i];
	cin >> t;
	rep(i, t){
		cin >> name[i] >> P[i];
	}
	vector<hito> d(4010);
	map<string, int> meibo;
	int right = 0;
	rep(i, t){
		if(meibo.count(name[i]) != 0){//いる
			int idx = meibo[name[i]];
			// printf("idx %d\n", idx);
			int num = P[i] - 'A';
			int hosi = L[num];
			int rank = cnt[P[i] - 'A'];
			cnt[P[i] - 'A']++;
			int tscore = 50.0 * (hosi) + (50 * (hosi)) / (0.8 + 0.2 * (rank + 1));
			// printf("%d\n", tscore);
			d[idx].score[num] = tscore;
			d[idx].sum += tscore;
			d[idx].kousin = i;
		}else{
			int idx = right;
			meibo.insert(make_pair(name[i], idx));

			right++;
			d[idx].na = name[i];
			int num = P[i] - 'A';
			int hosi = L[num];
			int rank = cnt[P[i] - 'A'];
			cnt[P[i] - 'A']++;
			int tscore = 50.0 * (hosi) + (50 * (hosi)) / (0.8 + 0.2 * (rank + 1));

			d[idx].score[num] = tscore;
			d[idx].sum += tscore;
			d[idx].kousin = i;
		}
	}
	/*
	rep(i, 2){
		printf("%d %s %d\n", i + 1, d[i].na, d[i].sum);
	}
	*/

	d.erase(d.begin() + right, d.end());
	sort(all(d), asc);
	rep(i, d.size()){
		printf("%d ", i + 1);
		cout << d[i].na << " ";
		rep(j, n){
			printf("%d ", d[i].score[j]);
		}
		printf("%d\n", d[i].sum);
	}
	return 0;
}