Tavas is a cheerleader in the new sports competition named "Pashmaks".

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This competition consists of two part: swimming and then running. People will immediately start running R meters after they finished swimming exactly S meters. A winner is a such person that nobody else finishes running before him/her (there may be more than one winner).

Before the match starts, Tavas knows that there are n competitors registered for the match. Also, he knows that i-th person's swimming speed is si meters per second and his/her running speed is ri meters per second. Unfortunately, he doesn't know the values of R and S, but he knows that they are real numbers greater than 0.

As a cheerleader, Tavas wants to know who to cheer up. So, he wants to know all people that might win. We consider a competitor might win if and only if there are some values of R and S such that with these values, (s)he will be a winner.

Tavas isn't really familiar with programming, so he asked you to help him.

Input

The first line of input contains a single integer n (1 ≤ n ≤ 2 × 105).

The next n lines contain the details of competitors. i-th line contains two integers si and ri (1 ≤ si, ri ≤ 104).

Output

In the first and the only line of output, print a sequence of numbers of possible winners in increasing order.

Examples

Input

3
1 3
2 2
3 1


Output

1 2 3 


Input

3
1 2
1 1
2 1


Output

1 3 
5
10 50
10 50
10 50
10 50
10 50
 2
10000 10000
10000 10000
 3
10000 10000
10000 10000
10000 10000
 3
1000 3000
1500 1500
3000 1000
 43
75 16
70 7
75 16
54 2
70 7
54 2
70 7
54 2
70 7
54 2
54 2
70 7
70 7
54 2
70 7
75 16
70 7
54 2
54 2
54 2
54 2
54 2
54 2
75 16
70 7
70 7
54 2
70 7
75 16
54 2
75 16
75 16
54 2
54 2
70 7
70 7
75 16
70 7
70 7
70 7
75 16
75 16
54 2
 1
1 1
 2
1 1
1 1
 44
78 52
75 16
70 7
75 16
54 2
70 7
54 2
70 7
54 2
70 7
54 2
54 2
70 7
70 7
54 2
70 7
75 16
70 7
54 2
54 2
54 2
54 2
54 2
54 2
75 16
70 7
70 7
54 2
70 7
75 16
54 2
75 16
75 16
54 2
54 2
70 7
70 7
75 16
70 7
70 7
70 7
75 16
75 16
54 2
 3
1000 3000
1499 1499
3000 1000
 5
50 50
49 50
50 49
49 49
50 1
 18
82 38
33 69
33 69
33 69
33 69
33 69
82 38
33 69
33 69
33 69
82 38
33 69
33 69
82 38
82 38
33 69
33 69
33 69
 1
10000 9999
 3
1000 3000
1501 1501
3000 1000
 2
1 2
2 1

1 2 3 4 5
 1 2
 1 2 3
 1 2 3
 1 3 16 24 29 31 32 37 41 42
 1
 1 2
 1
 1 3
 1
 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
 1
 1 2 3
 1 2 

#include<bits/stdc++.h>
using namespace std;

#define int long long 

#define fs first
#define sc second
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define sz(x) (int)x.size()
#define pb push_back
#define fastio ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define MP make_pair
#define sqr(x) ((x) * (x))
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

#define double long double

const double EPS = 1e-13;
struct vec {
	double x, y;
	int id;
	vec(){}
	vec(double x, double y, int id) : x(x), y(y), id(id) {}
	bool operator < (vec &o) {
		if (abs(x - o.x) < EPS)
			return y < o.y;
		return x < o.x;
	}
	double len() {
		return sqrt(sqr(x) + sqr(y));
	}
};

vec operator -(vec a, vec b) {return vec(a.x - b.x, a.y - b.y, -1);}
vec operator +(vec a, vec b) {return vec(a.x + b.x, a.y + b.y, -1);}

double dist(vec a, vec b) {
	return (a - b).len();
}

double operator ^(vec a, vec b) {
	return a.x * b.y - a.y * b.x;
}

ostream &operator << (ostream &out, vec &pt) {
	out << pt.x << ' ' << pt.y << '\n';
	return out;
}

vector<vec> ConvexHull(vector<vec> a) {
	swap(a[0], a[min_element(all(a)) - a.begin()]);

	sort(a.begin() + 1, a.end(), [&](vec i, vec j) {
		if (abs((i - a[0]) ^ (j - a[0])) < EPS)
			return dist(a[0], i) < dist(a[0], j);
		return ((i - a[0]) ^ (j - a[0])) > 0;
	});
	/*for (auto x : a) {
		cout << x;
	}*/
	vector<vec> res;
	for (auto p : a) {
		while (sz(res) > 1 && ((res.rbegin()[0] - res.rbegin()[1]) ^ (p - res.rbegin()[0])) <= -EPS) {
			res.pop_back();
		}
		res.pb(p);
	}
	return res;
}

signed main() {
	fastio;
	#ifdef FlameDragon
		freopen("in.txt", "r", stdin);
		freopen("out.txt", "w", stdout);
	#endif
	
	int n;
	cin >> n;
	vector<vec> a(n);
	for (int i = 0; i < n; i++) {
		int x, y;
		cin >> x >> y;
		a[i] = {1.0 / x, 1.0 / y, i};
	}
	a = ConvexHull(a);
	// cout << sz(a) << '\n';
	vec p = a[0];
	int nxt = 0;
	for (int i = 1; i < sz(a); i++) {
		if (abs(a[i].y - p.y) < EPS && a[i].x <= p.x + EPS || a[i].y < p.y) {
			nxt = i;
			p = a[i];
		}
	}
	// cout << nxt << endl << a[nxt];
	vector<int> res;
	for (int i = 0; i <= nxt; i++) {
		res.pb(a[i].id);
	}
	sort(all(res));
	for (auto x : res) {
		cout << x + 1 << ' ';
	}
	cout << '\n';
}