There are a total of N (1≤N≤5000) cows on the number line, each of which is a Holstein or a Guernsey. The breed of the i-th cow is given by bi∈H,G, the location of the i-th cow is given by xi (0≤xi≤109), and the weight of the i-th cow is given by yi (1≤yi≤105).
At Farmer John's signal, some of the cows will form pairs such that
- Every pair consists of a Holstein h and a Guernsey g whose locations are within K of each other (1≤K≤109); that is, |xh−xg|≤K.
- Every cow is either part of a single pair or not part of a pair.
- The pairing is maximal; that is, no two unpaired cows can form a pair.
It's up to you to determine the range of possible sums of weights of the unpaired cows. Specifically,
- If T=1, compute the minimum possible sum of weights of the unpaired cows.
- If T=2, compute the maximum possible sum of weights of the unpaired cows.
Input Format
The first input line contains T, N, and K.
Following this are N lines, the i-th of which contains bi,xi,yi. It is guaranteed that 0≤x1<x2<⋯<xN≤109.
Output
The minimum or maximum possible sum of weights of the unpaired cows.
Examples
Input 1
2 5 4 G 1 1 H 3 4 G 4 2 H 6 6 H 8 9
Output 1
16
Cows 2 and 3 can pair up because they are at distance 1, which is at most K=4. This pairing is maximal, because cow 1, the only remaining Guernsey, is at distance 5 from cow 4 and distance 7 from cow 5, which are more than K=4. The sum of weights of unpaired cows is 1+6+9=16.
Input 2
1 5 4
G 1 1
H 3 4
G 4 2
H 6 6
H 8 9
Output 2
6
Cows 1 and 2 can pair up because they are at distance 2≤K=4, and cows 3 and 5 can pair up because they are at distance 4≤K=4. This pairing is maximal because only cow 4 remains. The sum of weights of unpaired cows is the weight of the only unpaired cow, which is simply 6.
Input 3
2 10 76
H 1 18
H 18 465
H 25 278
H 30 291
H 36 202
G 45 96
G 60 375
G 93 941
G 96 870
G 98 540
Output 3
1893
The answer to this example is 18+465+870+540=1893.
Scoring
- Test cases 4-7 satisfy T=1.
- Test cases 8-14 satisfy T=2 and N≤300.
- Test cases 15-22 satisfy T=2.
Note: the memory limit for this problem is 512MB, twice the default.
Problem credits: Benjamin Qi