#include <bits/stdc++.h>
#define per pair<int,int>
using namespace std;
int last , n, m, aux[100005];
per a[100005];
inline bool solve(int timp)
{
    int ramase = m;
    for (int i = 1; i<=n && ramase > 0; ++i)
    {
        long long necesar = 2 * a[i].second;
        long long posibil = timp / necesar;
        ramase -= a[i].first * posibil;
    }
    return (ramase <= 0);
}
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 = INT_MAX - 1;
    while (st <= dr)
    {
        int mid = (st + dr) >> 1;
        if (solve(mid))
            last = mid , dr = mid - 1;
        else
            st = mid + 1;
    }
    printf("%d ", last);
    for (int i = 1; i<=n; ++i)
        aux[i] = 1LL * a[i].first * 1LL * last / 1LL / a[i].second / 2LL;
    sort(aux + 1,aux + n + 1);
    int ans = 0;
    for (int i = n; i>=1 && m > 0; --i)
    {
        m-=aux[i];
        ++ans;
    }
    printf("%d\n", ans);
    return 0;
}