/*input
3 40
10 1
10 2
10 3
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;
typedef vector<int> vi;
typedef vector<vi> vii;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef vector<pi> vpi;
typedef vector<vpi> vpii;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
typedef vector<pd> vpd;
typedef vector<bool> vb;
typedef vector<vb> vbb;
typedef std::string str;
typedef std::vector<str> vs;
#define x first
#define y second
#define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n"
const int MOD = 1000000007;
const ll INF = std::numeric_limits<ll>::max();
const int MX = 100101;
const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L;
template<typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x + b.x, a.y + b.y); }
template<typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x - b.x, a.y - b.y); }
template<typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.x + a.y * b.y); }
template<typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.y - a.y * b.x); }
template<typename T>
void print(vector<T> vec, string name = "") {
cout << name;
for (auto u : vec)
cout << u << ' ';
cout << '\n';
}
struct ppl {
ll S, Q;
int i;
};
int main() {
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
int N;
ll W;
cin >> N >> W;
vector<ppl> sk(N);
for (int i = 0; i < N; ++i)
{
int a, b; cin >> a >> b;
sk[i] = {a, b, i + 1};
}
sort(sk.begin(), sk.end(), [&](const ppl & a, const ppl & b) {
return a.Q * b.S < b.Q * a.S;
});
ll sumQ = 0;
int kiek = 0;
int l = 1, r = 0;
ll sumokejau = MOD;
int j = 1;
sumQ = sk[0].Q;
for (int i = 0; i < N; ++i)
{
while (j < N and (sumQ + sk[j].Q)*sk[i].S <= W * sk[i].Q) {
sumQ += sk[j].Q;
j++;
}
if (sumQ * sk[i].S <= W * sk[i].Q) {
if (j - i > kiek or (j - i == kiek and sumQ * sk[i].S * sk[l].Q < sumokejau * sk[i].Q)) {
l = i;
r = j - 1;
kiek = j - i;
sumokejau = sumQ * sk[i].S;
}
}
sumQ -= sk[i].Q;
}
printf("%d\n", kiek);
for (int i = l; i <= r; ++i)
{
printf("%d\n", sk[i].i);
}
}
/* Look for:
* special cases (n=1?)
* overflow (ll vs int?)
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* array bounds
*/
