#include <bits/stdc++.h>
#define per pair<int,int>
using namespace std;
int last , n, m;
per a[100005];
inline bool solve(int timp)
{
    int ramase = m;
    for (int i = 1; i<=n && ramase > 0; ++i)
    {
        int necesar = 2 * a[i].second;
        int posibil = timp / necesar;
        ramase -= a[i].first * posibil;
    }
    return (ramase <= 0);
}
inline bool cmp(per a, per b)
{
    int necesar1 = 2 * a.second;
    int posibil1 = last / necesar1;
    int carate1 = a.first * posibil1;
    int necesar2 = 2 * b.second;
    int posibil2 = last / necesar2;
    int carate2 = b.first * posibil2;
    return carate1 > carate2;
}
int main()
{
    freopen("garaj.in", "r", stdin);
    freopen("garaj.out", "w", stdout);

    scanf("%d %d\n", &n, &m);
    for (int i = 1; i<=n; ++i)
        scanf("%d %d\n", &a[i].first, &a[i].second);
    int st = 1, dr = 1e5;
    while (st <= dr)
    {
        int mid = (st + dr) >> 1;
        if (solve(mid))
            last = mid , dr = mid - 1;
        else
            st = mid + 1;
    }
    printf("%d ", last);
    sort(a + 1, a + n + 1, cmp);
    int ans = 0;
    for (int i = 1; i<=n && m > 0; ++i)
    {
        int necesar = 2 * a[i].second;
        int posibil = last / necesar;
        m -= a[i].first * posibil;
        ++ans;
    }
    printf("%d\n", ans);
    return 0;
}