//朴素解法
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 110;

int v[N][N], w[N][N], s[N]; //v为体积，w为价值，s代表第i组物品的个数
int dp[N][N];               //只从前i组物品中选，当前体积小于等于j的最大值

int main()
{
    int n, V;
    cin >> n >> V;
    for(int i = 1; i <= n; i++)
    {
        cin >> s[i];
        for(int j = 1; j <= s[i]; j++)
        {
            cin >> v[i][j] >> w[i][j];  //读入
        }
    }

    for(int i = 1; i <= n; i++)
    {
        for(int j = 0; j <= V; j++)
        {
            dp[i][j] = dp[i - 1][j];    //不选
            for(int k = 1; k <= s[i]; k++)
            {
                if(v[i][k] <= j) dp[i][j] = max(dp[i][j], dp[i - 1][j - v[i][k]] + w[i][k]);
            }
        }
    }

    cout << dp[n][V] << endl;

    return 0;
}

//优化解法
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 110;

int n, m;
int v[N][N], w[N][N], s[N];
int f[N];

int main()
{
    cin >> n >> m;

    for(int i = 1; i <= n; i++)
    {
        cin >> s[i];
        for(int j = 0; j < s[i]; j++)
        {
            cin >> v[i][j] >> w[i][j];
        }
    }

    for(int i = 1; i <= n; i++)
    {
        for(int j = m; j >= 0; j--)
        {
            for(int k = 0; k < s[i]; k++)
            {
                if(v[i][k] <= j) f[j] = max(f[j], f[j - v[i][k]] + w[i][k]);
            }
        }
    }

    cout << f[m] << endl;

    return 0;
}