Problem: You are given a set of n≥ 2 pairwise different points with integer coordinates. Your task is to partition these points into two nonempty groups A and B, such that the following condition holds:

For every two points P and Q, write the [Euclidean distance](https://en.wikipedia.org/wiki/Euclidean_distance) between them on the blackboard: if they belong to the same group — with a yellow pen, and if they belong to different groups — with a blue pen. Then no yellow number is equal to any blue number.

It is guaranteed that such a partition exists for any possible input. If there exist multiple partitions, you are allowed to output any of them.

Input

The first line contains one integer n (2 ≤ n ≤ 10^3) — the number of points.

The i-th of the next n lines contains two integers x_i and y_i (-10^6 ≤ x_i, y_i ≤ 10^6) — the coordinates of the i-th point. 

It is guaranteed that all n points are pairwise different.

Output

In the first line, output a (1 ≤ a ≤ n-1) — the number of points in a group A.

In the second line, output a integers — the indexes of points that you include into group A.

If there are multiple answers, print any.

Examples

Input


3
0 0
0 1
1 0


Output


1
1 

Input


4
0 1
0 -1
1 0
-1 0


Output


2
1 2 

Input


3
-2 1
1 1
-1 0


Output


1
2 

Input


6
2 5
0 3
-4 -1
-5 -4
1 0
3 -1


Output


1
6 

Input


2
-1000000 -1000000
1000000 1000000


Output


1
1 

Note

In the first example, we set point (0, 0) to group A and points (0, 1) and (1, 0) to group B. In this way, we will have 1 yellow number √{2} and 2 blue numbers 1 on the blackboard.

In the second example, we set points (0, 1) and (0, -1) to group A and points (-1, 0) and (1, 0) to group B. In this way, we will have 2 yellow numbers 2, 4 blue numbers √{2} on the blackboard.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
ostream& operator<<(ostream& os, vector<T> V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  return os << "]";
}
template <class T>
ostream& operator<<(ostream& os, set<T> second) {
  os << "{ ";
  for (auto s : second) os << s << " ";
  return os << "}";
}
template <class L, class R>
ostream& operator<<(ostream& os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
template <class L, class R>
ostream& operator<<(ostream& os, map<L, R> M) {
  os << "{ ";
  for (auto m : M) os << "(" << m.first << ":" << m.second << ") ";
  return os << "}";
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const long long maxn = 2e6 + 100;
const long long MaxN = 1e5 + 100;
const long long mod = 1e9 + 7;
const long long INF = 1e16;
inline long long fast_expo(long long base, long long power,
                           long long modulo = mod) {
  base %= modulo;
  if (base < 0) base += modulo;
  long long x = base, cnt = power, ans = 1;
  while (cnt) {
    if (cnt & 1) ans = (ans * x) % modulo;
    x = (x * x) % modulo;
    cnt >>= 1;
  }
  return ans;
}
inline long long inv(long long base, long long modulo = mod) {
  return fast_expo(base, modulo - 2, modulo);
}
long long x[maxn], y[maxn];
const long long N = 1040;
long long d[N][N];
inline long long dis(long long i, long long j) {
  if (d[i][j] == 0)
    d[i][j] = d[j][i] =
        ((x[i] - x[j]) * (x[i] - x[j])) + ((y[i] - y[j]) * (y[i] - y[j]));
  return d[i][j];
}
long long dsu[maxn];
long long sz[maxn];
set<long long> G[maxn];
inline long long Fi(long long x) {
  return (x == dsu[x]) ? x : dsu[x] = Fi(dsu[x]);
}
inline void Me(long long x, long long y) {
  x = Fi(x);
  y = Fi(y);
  if (x == y) return;
  if (sz[x] < sz[y]) swap(x, y);
  dsu[y] = x;
  sz[x] += sz[y];
}
map<long long, long long> cnt;
set<pair<long long, long long> > Pos[maxn];
bool sam[maxn];
vector<long long> adj[maxn];
map<long long, long long> idx;
void solve() {
  long long n;
  cin >> n;
  for (long long i = 0; i < (long long)n; i++) {
    cin >> x[i] >> y[i];
  }
  for (long long i = 0; i < (long long)n; i++) {
    for (long long j = i + 1; j < n; ++j) ++cnt[dis(i, j)];
  }
  long long us = 0;
  for (auto e : cnt) {
    idx[e.first] = ++us;
  }
  for (long long i = 0; i < (long long)n; i++) {
    unordered_map<long long, long long> res;
    for (long long j = 0; j < (long long)n; j++) {
      if (i != j) {
        ++res[dis(i, j)];
      }
    }
    bool pos = 1;
    for (auto e : res) {
      if (e.second != cnt[e.first]) {
        pos = 0;
        break;
      }
    }
    if (pos) {
      cout << 1 << "\n";
      cout << i + 1 << "\n";
      return;
    }
  }
  for (long long i = 0; i < (long long)n; i++) {
    dsu[i] = i;
    sz[i] = 1;
  }
  for (long long i = 0; i < (long long)n; i++) {
    unordered_map<long long, long long> prv;
    for (long long j = 0; j < (long long)n; j++) {
      if (i != j) {
        long long z = dis(i, j);
        if (prv.find(z) != prv.end()) {
          Me(j, prv[z]);
        }
        prv[z] = j;
      }
    }
  }
  for (long long i = 0; i < (long long)n; i++) {
    adj[Fi(i)].push_back(i);
  }
  for (long long i = 0; i < (long long)n; i++) {
    for (long long j = i + 1; j < n; ++j) {
      Pos[idx[dis(i, j)]].insert({min(Fi(i), Fi(j)), max(Fi(i), Fi(j))});
      if (Fi(i) == Fi(j)) {
        sam[idx[dis(i, j)]] = 1;
      }
    }
  }
  for (long long i = 0; i < (long long)2 * n; i++) {
    sz[i] = 1;
    dsu[i] = i;
  }
  for (long long i = 0; i < (long long)n; i++) {
    for (auto e : adj[i]) {
      Me(2 * i, 2 * e);
      Me(2 * i + 1, 2 * e + 1);
    }
  }
  for (long long i = 1; i <= us; ++i) {
    if (sam[i]) {
      for (auto e : Pos[i]) {
        Me(2 * e.first, 2 * e.second);
        Me(2 * e.first + 1, 2 * e.second + 1);
      }
    }
  }
  for (long long i = 1; i <= us; ++i) {
    if (!sam[i]) {
      bool op = 1;
      for (auto e : Pos[i]) {
        long long s1 = Fi(2 * e.first), s2 = Fi(2 * e.second);
        if (s1 == s2) op = 0;
      }
      if (op) {
        for (auto e : Pos[i]) {
          Me(2 * e.first, 2 * e.second + 1);
          Me(2 * e.first + 1, 2 * e.second);
        }
      }
    }
  }
  vector<long long> A;
  A.push_back(0);
  long long unuse = -1;
  long long prv = 0;
  vector<long long> B;
  vector<long long> rem;
  for (long long i = 1; i < n; ++i) rem.push_back(i);
  while (!rem.empty() && prv >= 0) {
    vector<long long> tp;
    for (auto j : rem) {
      if (Fi(2 * j) == Fi(2 * prv)) {
        A.push_back(j);
      } else if (Fi(2 * j + 1) == Fi(2 * prv)) {
        B.push_back(j);
      } else {
        if (unuse != -1) tp.push_back(unuse);
        unuse = j;
      }
    }
    rem.swap(tp);
    A.swap(B);
    prv = unuse;
    unuse = -1;
    if (prv >= 0) A.push_back(prv);
  }
  cout << A.size() << "\n";
  for (auto e : A) cout << e + 1 << " ";
  cout << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  while (t--) {
    solve();
  }
}