1 // state expression:
 2 // dp[i][j][k] = max value
 3 // i: considering ith DVD
 4 // j: j DVDs is selected
 5 // k: cost time
 6 //
 7 // find the answer:
 8 // max dp[n][m][0 to l]
 9 //
10 // transferring:
11 // now: dp[i][j][k]
12 // dp[i+1][j][k] = max dp[i][j][k]
13 // dp[i+1][j+1][k+t[i]] = max dp[i][j][k] + w[i]
14 //
15 // boundary:
16 // dp[0][0][0] = 0
17 // others = -1
18 #include <iostream>
19 #include <stdio.h>
20 #include <string.h>
21 #define MAX_N 105
22 #define MAX_M 105
23 #define MAX_L 1005
24
25 using namespace std;
26
27 int n,m,l;
28 int ans;
29 int t[MAX_N];
30 int w[MAX_N];
31 int dp[MAX_N][MAX_M][MAX_L];
32
33 void read()
34 {
35     cin>>n>>m>>l;
36     for(int i=0;i<n;i++)
37     {
38         cin>>t[i]>>w[i];
39     }
40 }
41
42 void solve()
43 {
44     memset(dp,-1,sizeof(dp));
45     dp[0][0][0]=0;
46     for(int i=0;i<n;i++)
47     {
48         for(int j=0;j<=m;j++)
49         {
50             for(int k=0;k<=l;k++)
51             {
52                 if(dp[i][j][k]!=-1)
53                 {
54                     dp[i+1][j][k]=max(dp[i+1][j][k],dp[i][j][k]);
55                     if(j+1<=m && k+t[i]<=l)
56                     {
57                         dp[i+1][j+1][k+t[i]]=max(dp[i+1][j+1][k+t[i]],dp[i][j][k]+w[i]);
58                     }
59                 }
60             }
61         }
62     }
63     ans=0;
64     for(int i=0;i<=l;i++)
65     {
66         ans=max(ans,dp[n][m][i]);
67     }
68 }
69
70 void print()
71 {
72     cout<<ans<<endl;
73 }
74
75 int main()
76 {
77     read();
78     solve();
79     print();
80 }