import sys

def max_points(n, T, m, items):
    # dp[j][k]：用 j 个超长用例、时间 k 的最大测试点，不可达为 -1
    dp = [[-1] * (T + 1) for _ in range(m + 1)]
    dp[0][0] = 0  # 什么都不选

    for w, v in items:
        long_cnt = 1 if w > 30 else 0  # 是否为超长用例
        # 倒序枚举，保证 0/1（每个用例最多选一次）
        for j in range(m, long_cnt - 1, -1):
            row = dp[j]
            prev = dp[j - long_cnt]
            for k in range(T, w - 1, -1):
                if prev[k - w] != -1:
                    cand = prev[k - w] + v
                    if cand > row[k]:
                        row[k] = cand

    ans = 0
    for j in range(m + 1):
        best = max(dp[j])
        if best > ans:
            ans = best
    return ans

def main():
    data = sys.stdin.read().strip().split()
    if not data:
        return
    it = iter(data)
    n = int(next(it))
    T = int(next(it))
    m = int(next(it))
    items = []
    for _ in range(n):
        w = int(next(it))
        v = int(next(it))
        items.append((w, v))
    print(max_points(n, T, m, items))

if __name__ == "__main__":
    main()

#include <bits/stdc++.h>
using namespace std;
// 题面功能：带“超长用例数量限制”的 0/1 背包
int maxPoints(int n, int T, int m, const vector<pair<int,int>>& items) {
    // dp[j][k]：用 j 个超长用例、时间 k 的最大测试点，不可达为 -1
    vector<vector<int>> dp(m + 1, vector<int>(T + 1, -1));
    dp[0][0] = 0; // 什么都不选

    for (auto [w, v] : items) {
        int longCnt = (w > 30) ? 1 : 0; // 是否为超长用例
        // 倒序枚举，保证 0/1（每个用例最多选一次）
        for (int j = m; j >= longCnt; --j) {
            for (int k = T; k >= w; --k) {
                if (dp[j - longCnt][k - w] != -1) {
                    dp[j][k] = max(dp[j][k], dp[j - longCnt][k - w] + v);
                }
            }
        }
    }
    int ans = 0;
    for (int j = 0; j <= m; ++j) {
        for (int k = 0; k <= T; ++k) {
            ans = max(ans, dp[j][k]);
        }
    }
    return ans;
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, T, m;
    cin >> n >> T >> m;
    vector<pair<int,int>> items(n);
    for (int i = 0; i < n; ++i) {
        int t, p;
        cin >> t >> p;
        items[i] = {t, p};
    }

    cout << maxPoints(n, T, m, items) << "\n";
    return 0;
}

import java.io.*;
import java.util.*;

public class Main {

    // 题面功能：带“超长用例数量限制”的 0/1 背包
    static int maxPoints(int n, int T, int m, int[] w, int[] v) {
        // dp[j][k]：用 j 个超长用例、时间 k 的最大测试点，不可达为 -1
        int[][] dp = new int[m + 1][T + 1];
        for (int j = 0; j <= m; j++) Arrays.fill(dp[j], -1);
        dp[0][0] = 0; // 什么都不选

        for (int i = 0; i < n; i++) {
            int time = w[i];
            int points = v[i];
            int longCnt = (time > 30) ? 1 : 0; // 是否超长

            // 倒序枚举，保证每个用例最多选一次
            for (int j = m; j >= longCnt; j--) {
                for (int k = T; k >= time; k--) {
                    if (dp[j - longCnt][k - time] != -1) {
                        int cand = dp[j - longCnt][k - time] + points;
                        if (cand > dp[j][k]) dp[j][k] = cand;
                    }
                }
            }
        }

        int ans = 0;
        for (int j = 0; j <= m; j++) {
            for (int k = 0; k <= T; k++) {
                if (dp[j][k] > ans) ans = dp[j][k];
            }
        }
        return ans;
    }

    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st;

        String first = br.readLine();
        if (first == null || first.trim().isEmpty()) return;
        st = new StringTokenizer(first);
        int n = Integer.parseInt(st.nextToken());
        int T = Integer.parseInt(st.nextToken());
        int m = Integer.parseInt(st.nextToken());

        int[] w = new int[n];
        int[] v = new int[n];
        for (int i = 0; i < n; i++) {
            st = new StringTokenizer(br.readLine());
            w[i] = Integer.parseInt(st.nextToken());
            v[i] = Integer.parseInt(st.nextToken());
        }

        System.out.println(maxPoints(n, T, m, w, v));
    }
}