/*#include <fstream>
using namespace std;

const int maxn = 5002;
const int maxg = 10002;

int n, g, w[maxn], p[maxn], M[maxn][maxg];

int main()
{
    ifstream fin("rucsac.in");
    ofstream fout("rucsac.out");

    fin>>n>>g;
    for(int i=1; i<=n; i++)
    {
        fin>>w[i]>>p[i];
    }
    for(int i=1; i<=n; i++)
    {
        for(int j=0; j<=g; j++)
        {
            if(w[i]>j)
            {
                M[i][j]=M[i-1][j];
            }
            else
            {
                M[i][j]=max(M[i-1][j],M[i-1][j-w[i]]+p[i]);
            }
        }
    }
    fout<<M[n][g]<<'\n';
    return 0;
}*/

#include <cstdio>
#include <algorithm>

using namespace std;

#define MAXN 5010
#define MAXG 10010

int N, G, Pmax;
int W[MAXN], P[MAXN];
int D[MAXN][MAXG];

int main()
{
    freopen("rucsac.in", "r", stdin);
    freopen("rucsac.out", "w", stdout);

    // Citire
    scanf("%d%d", &N, &G);
    for(int i = 1; i <= N; ++i)
        scanf("%d%d", &W[i], &P[i]);

    // Dinamica D[i][cw] - profitul maxim pe care-l putem obtine adaugand o submultime a primelor i obiecte, insumand greutatea cw
    for(int i = 1; i <= N; ++i)
        for(int cw = 0; cw <= G; ++cw)
        {
            // Mai intai nu punem obiectul i.
            D[i][cw] = D[i-1][cw];

            // Daca acest lucru duce la o solutie curenta mai buna, adaugam obiectul i la o solutie anterioara.
            if(W[i] <= cw)
                D[i][cw] = max(D[i][cw], D[i - 1][cw - W[i]] + P[i]);
        }

    // Solutia se va afla in statea D[N][G]
    Pmax = D[N][G];

    // Afisare
    printf("%d\n", Pmax);

    return 0;
}