The first line contains three integers n, c and d (2?≤?n?≤?100?000, 0?≤?c,?d?≤?100?000) — the number of fountains, the number of coins and diamonds Arkady has.

The next n lines describe fountains. Each of these lines contain two integers b_i and p_i (1?≤?b_i,?p_i?≤?100?000) — the beauty and the cost of the i-th fountain, and then a letter "C" or "D", describing in which type of money is the cost of fountain i: in coins or in diamonds, respectively.

Print the maximum total beauty of exactly two fountains Arkady can build. If he can‘t build two fountains, print 0.

In the first example Arkady should build the second fountain with beauty 4, which costs 3 coins. The first fountain he can‘t build because he don‘t have enough coins. Also Arkady should build the third fountain with beauty 5 which costs 6 diamonds. Thus the total beauty of built fountains is 9.

In the second example there are two fountains, but Arkady can‘t build both of them, because he needs 5 coins for the first fountain, and Arkady has only 4 coins.

 1 #include <iostream>
 2 #include <string>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <cstdio>
 6 #include <cmath>
 7 #include <queue>
 8 #include <set>
 9 #include <map>
10 #include <list>
11 #include <stack>
12 #define mp make_pair
13 typedef long long ll;
14 typedef unsigned long long ull;
15 const int MAX=1e5;
16 const int INF=1e9+5;
17 const double M=4e18;
18 using namespace std;
19 const int MOD=1e9+7;
20 typedef pair<int,int> pii;
21 const double eps=0.000000001;
22 int C[MAX+5],D[MAX+5];
23 void add(int *tree,int k,int num)
24 {
25     while(k<=MAX)
26     {
27         tree[k]=max(tree[k],num);
28         k+=k&(-k);
29     }
30 }
31 int read(int *tree,int k)
32 {
33     int re=0;
34     while(k)
35     {
36         re=max(re,tree[k]);
37         k-=k&(-k);
38     }
39     return re;
40 }
41 int n,c,d,b,p;
42 int an,tem;
43 char opt[10];
44 int main()
45 {
46     scanf("%d%d%d",&n,&c,&d);
47     while(n--)
48     {
49         scanf("%d%d%s",&b,&p,opt);
50         tem=0;
51         if(opt[0]==‘C‘)
52         {
53             if(p>c)
54                 continue;
55             tem=read(D,d);
56             tem=max(tem,read(C,c-p));
57             add(C,p,b);
58         }
59         else
60         {
61             if(p>d)
62                 continue;
63             tem=read(C,c);
64             tem=max(tem,read(D,d-p));
65             add(D,p,b);
66         }
67         if(tem)
68             an=max(an,tem+b);
69     }
70     printf("%d\n",an);
71     return 0;
72 }