// 题意:输入n个矩阵的维度和一些矩阵链乘表达式,输出乘法的次数。假定A和m*n的,B是n*p的,那么AB是m*p的,乘法次数为m*n*p
// 算法:用一个栈。遇到字母时入栈,右括号时出栈并计算,然后结果入栈。因为输入保证合法,括号无序入栈
// Created by Chenhongwei in 2015.
// Copyright (c) 2015 Chenhongwei. All rights reserved.
#include"iostream"
#include"cstdio"
#include"cstdlib"
#include"cstring"
#include"climits"
#include"queue"
#include"cmath"
#include"map"
#include"set"
#include"vector"
#include"stack"
#include"sstream"
#include"algorithm"
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
P m[30];
int main()
{
//ios::sync_with_stdio(false);
// freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n;
scanf("%d\n",&n);
for(int i=1;i<=n;i++)
{
char ch;
int a,b;
scanf("%c %d %d\n",&ch,&a,&b);
m[ch-'A'+1].first=a;
m[ch-'A'+1].second=b;
}
string s;
while(getline(cin,s))
{
stack<P> stk;
ll ans=0;
int flag=1;
for(int i=0;i<s.size();i++)
{
if(s[i]=='(')continue;
else if(isalpha(s[i]))
stk.push(m[s[i]-'A'+1]);
else if(s[i]==')')
{
P x,y,temp;
x=stk.top();stk.pop();
y=stk.top();stk.pop();
if(x.first!=y.second)
{
flag=0;
printf("error\n");
break;
}
else
{
ans+=y.first*y.second*x.second;
temp.first=y.first;
temp.second=x.second;
stk.push(temp);
}
}
}
if(flag==1)
cout<<ans<<endl;
}
return 0;
}