#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int mxN=2e5;
int n, l, r;
ll x[mxN], y[mxN];
vector<int> v, s;
bool b[mxN];
ll cp(int a, int b, int c) {
return (y[c]-y[a])*(x[b]-x[a])-(y[b]-y[a])*(x[c]-x[a]);
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cin >> n;
for(int i=0; i<n; ++i) {
cin >> x[i] >> y[i];
if(!i||cp(n, l, i)>0||!cp(n, l, i)&&x[i]<x[l])
l=i;
if(!i||cp(n, r, i)<0||!cp(n, r, i)&&x[i]<x[r])
r=i;
}
//cout << l << " " << r << endl;
int ic=cp((l+n-1)%n, l, (l+1)%n)>0?1:n-1;
for(; l^r; l=(l+ic)%n)
v.push_back(l);
v.push_back(r);
//cout << "a" << endl;
//for(int i : v)
//cout << x[i] << " " << y[i] << endl;
s={0, 1};
for(int i=2; i<v.size(); ++i) {
//cout << i << endl;
//for(int si : s)
//cout << si << " ";
//cout << endl;
//side going strictly forwards
if(cp(n, v[i-1], v[i])<0) {
//last side went strictly backwards and current side is now hidden
if(cp(n, v[i-2], v[i-1])>0&&cp(v[i-2], v[i-1], v[i])<0) {
//while hidden
while(cp(n, v[s.back()], v[i])<=0)
++i;
s.pop_back();
//while one of s.back() and i hides the other and it is actually s.back() which is hidden
while(cp(n, v[s.back()], v[i])>=0&&cp(v[i-1], v[i], v[s.back()])<0)
s.pop_back();
//i is hidden
if(cp(n, v[s.back()], v[i])>=0) {
int j=i, c=0;
//while i is not to the right of s.back() or i is hidden by segment j
while(cp(n, v[s.back()], v[i])>=0||!c) {
//if this is the first side, c will be one iff the new edge goes forwards and is not hidden by segment j
if(i==j)
c=cp(n, v[i], v[i+1])<0&&cp(v[i-1], v[i], v[i+1])>0;
//if we start not to the right of j but end up to the right of j and j does not hide i+1
if(cp(n, v[j], v[i])>=0&&cp(n, v[j], v[i+1])<0&&cp(v[i], v[i+1], v[j])>0)
++c;
//if we start to the right of j but end up not to the right of j and j does not hide i+1
if(cp(n, v[j], v[i])<0&&cp(n, v[j], v[i+1])>=0&&cp(v[i], v[i+1], v[j])<0)
--c;
++i;
while(i>=v.size());
}
}
}
}
//side going strictly backwards
else if(cp(n, v[i-1], v[i])>0) {
//last side went strictly forwards and current side is now hidden
if(cp(n, v[i-2], v[i-1])<0&&cp(v[i-2], v[i-1], v[i])>0) {
//while hidden
while(cp(n, v[s.back()], v[i])>=0)
++i;
} else {
//while one of s.back() and i hides the other and it is actually s.back() which is hidden
//note <= here so we can pop off v[i-1]
while(cp(n, v[s.back()], v[i])>=0&&cp(v[i-1], v[i], v[s.back()])<=0)
s.pop_back();
//i is hidden
if(cp(n, v[s.back()], v[i])>=0) {
int j=i, c=0;
//while i is not to the right of s.back() or i is hidden by segment j
while(cp(n, v[s.back()], v[i])>=0||!c) {
//if this is the first side, c will be one iff the new edge goes forwards and is not hidden by segment j
if(i==j)
c=cp(n, v[i], v[i+1])<0&&cp(v[i-1], v[i], v[i+1])>0;
//if we start not to the right of j but end up to the right of j and j does not hide i+1
if(cp(n, v[j], v[i])>=0&&cp(n, v[j], v[i+1])<0&&cp(v[i], v[i+1], v[j])>0)
++c;
//if we start to the right of j but end up not to the right of j and j does not hide i+1
if(cp(n, v[j], v[i])<0&&cp(n, v[j], v[i+1])>=0&&cp(v[i], v[i+1], v[j])<0)
--c;
++i;
while(i>=v.size());
}
}
}
}
//side not changing angle
else {
//if new point hides old point
if((cp(n, v[i-2], v[i-1])<0)==(cp(v[i-2], v[i-1], v[i])<0))
s.pop_back();
//last side went strictly forwards and current side is now hidden
else if(cp(n, v[i-2], v[i-1])<0) {
//while hidden
while(cp(n, v[s.back()], v[i])>=0)
++i;
}
//last side went strictly backwards and current side is now hidden
else {
//while hidden
while(cp(n, v[s.back()], v[i])<=0)
++i;
s.pop_back();
//while one of s.back() and i hides the other and it is actually s.back() which is hidden
while(cp(n, v[s.back()], v[i])>=0&&cp(v[i-1], v[i], v[s.back()])<0)
s.pop_back();
//i is hidden
if(cp(n, v[s.back()], v[i])>=0) {
int j=i, c=0;
//while i is not to the right of s.back() or i is hidden by segment j
while(cp(n, v[s.back()], v[i])>=0||!c) {
//if this is the first side, c will be one iff the new edge goes forwards and is not hidden by segment j
if(i==j)
c=cp(n, v[i], v[i+1])<0&&cp(v[i-1], v[i], v[i+1])>0;
//if we start not to the right of j but end up to the right of j and j does not hide i+1
if(cp(n, v[j], v[i])>=0&&cp(n, v[j], v[i+1])<0&&cp(v[i], v[i+1], v[j])>0)
++c;
//if we start to the right of j but end up not to the right of j and j does not hide i+1
if(cp(n, v[j], v[i])<0&&cp(n, v[j], v[i+1])>=0&&cp(v[i], v[i+1], v[j])<0)
--c;
++i;
while(i>=v.size());
}
}
}
}
s.push_back(i);
}
cout << s.size() << "\n";
for(int a : s)
b[v[a]]=1;
for(int i=0; i<n; ++i)
if(b[i])
cout << i+1 << " ";
}
