1 #include <set>
  2 #include <map>
  3 #include <queue>
  4 #include <stack>
  5 #include <cmath>
  6 #include <ctime>
  7 #include <bitset>
  8 #include <cstdio>
  9 #include <string>
 10 #include <vector>
 11 #include <cstdlib>
 12 #include <cstring>
 13 #include <iostream>
 14 #include <algorithm>
 15 using namespace std;
 16
 17 typedef long long ll;
 18 typedef pair<ll, ll> pll;
 19 typedef pair<ll, int> pli;
 20 typedef pair<int, ll> pil;;
 21 typedef pair<int, int> pii;
 22 typedef unsigned long long ull;
 23
 24 #define lson i<<1
 25 #define rson i<<1|1
 26 #define bug printf("*********\n");
 27 #define FIN freopen("D://code//in.txt", "r", stdin);
 28 #define debug(x) cout<<"["<<x<<"]" <<endl;
 29 #define IO ios::sync_with_stdio(false),cin.tie(0);
 30
 31 const double eps = 1e-8;
 32 const int mod = 1000000007;
 33 const int maxn = 1e5 + 7;
 34 const double pi = acos(-1);
 35 const int inf = 0x3f3f3f3f;
 36 const ll INF = 0x3f3f3f3f3f3f3f;
 37
 38 int n, c, d, bea, cost, mx1, mx2, t1, t2;
 39 char op[6];
 40
 41 struct node {
 42     int bea, cost;
 43     node(int bea = 0, int  cost = 0) : bea(bea), cost(cost) {}
 44 }num1[maxn], num2[maxn];
 45
 46 struct tree {
 47     int l, r, mx;
 48 }segtree[maxn*4];
 49
 50 void push_up(int i) {
 51     segtree[i].mx = max(segtree[i*2].mx, segtree[i*2+1].mx);
 52 }
 53
 54 void build(int i, int l, int r) {
 55     segtree[i].l = l, segtree[i].r = r;
 56     if(l == r) {
 57         segtree[i].mx = 0;
 58         return;
 59     }
 60     int mid = (l + r) >> 1;
 61     build(i * 2, l, mid);
 62     build(i * 2 + 1, mid + 1, r);
 63     push_up(i);
 64 }
 65
 66 void update(int i, int pos, int val) {
 67     if(segtree[i].l == pos && segtree[i].r == pos) {
 68         segtree[i].mx = max(segtree[i].mx, val);
 69         return;
 70     }
 71     int mid = (segtree[i].l + segtree[i].r) >> 1;
 72     if(pos <= mid) update(i*2, pos, val);
 73     else update(i*2+1, pos, val);
 74     push_up(i);
 75 }
 76
 77 int query(int i, int l, int r) {
 78     if(segtree[i].l == l && segtree[i].r == r) {
 79         return segtree[i].mx;
 80     }
 81     int mid = (segtree[i].l + segtree[i].r) >> 1;
 82     if(r <= mid) return query(i*2, l, r);
 83     else if(l > mid) return query(i*2+1, l, r);
 84     else return max(query(i*2, l, mid), query(i*2+1, mid + 1, r));
 85 }
 86
 87 int main() {
 88     scanf("%d%d%d", &n, &c, &d);
 89     mx1 = -1, mx2 = -1;
 90     for(int i = 1; i <= n; i++) {
 91         scanf("%d%d%s", &bea, &cost, op);
 92         if(op[0] == 'C') {
 93             if(cost <= c && bea > mx1) {
 94                 mx1 = bea;
 95             }
 96             num1[++t1] = node(bea, cost);
 97         } else {
 98             if(cost <= d && bea > mx2) {
 99                 mx2 = bea;
100             }
101             num2[++t2] = node(bea, cost);
102         }
103     }
104     int ans = 0;
105     build(1, 0, maxn); //因为C-num[i].cost可能等于0，所以线段树的区间需要从0开始
106     for(int i = 1; i <= t1; i++) {
107         if(c >= num1[i].cost) {
108             int t = query(1, 0, c - num1[i].cost);
109             if(t != 0) {
110                 ans = max(ans, t + num1[i].bea);
111             }
112         }
113         update(1, num1[i].cost, num1[i].bea);
114     }
115     build(1, 0, maxn);
116     for(int i = 1; i <= t2; i++) {
117         if(d >= num2[i].cost) {
118             int t = query(1, 0, d - num2[i].cost);
119             if(t != 0) {
120                 ans = max(ans, t + num2[i].bea);
121             }
122         }
123         update(1, num2[i].cost, num2[i].bea);
124     }
125     if(mx1 != -1 && mx2 != -1) { //一个用钻石支付，一个用现金支付
126         ans = max(ans, mx1 + mx2);
127     }
128     printf("%d\n", ans);
129     return 0;
130 }