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 

import java.io.IOException;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.Map;

public class Main {

    private static final double EPS = 1e-20;
    private static final int MAX = 10000;

    public static void main(String[] args) {
	int N = nextInt();
	HashMap<Integer, ArrayList<Integer>> map = new HashMap<>();

	int maxx = 0;
	int maxy = 0;
	for (int i = 0; i < N; i++) {
	    int x = nextInt() - 1;
	    int y = nextInt() - 1;
	    maxx = Math.max(maxx, x);
	    maxy = Math.max(maxy, y);

	    int hash = x * MAX + y;
	    if (map.get(hash) == null) {
		map.put(hash, new ArrayList<Integer>());
	    }
	    map.get(hash).add(i + 1);
	}

	if (map.get(0) == null) {
	    map.put(0, new ArrayList<Integer>());
	    map.get(0).add(N + 1);
	}
	if (map.get(maxx * MAX) == null) {
	    map.put(maxx * MAX, new ArrayList<Integer>());
	    map.get(maxx * MAX).add(N + 1);
	}
	if (map.get(maxy) == null) {
	    map.put(maxy, new ArrayList<Integer>());
	    map.get(maxy).add(N + 1);
	}

	ArrayList<Integer> ansList = new ArrayList<>();
	ArrayList<Integer> keyList = new ArrayList<>();
	int startKey = 0;
	// x座標のもっとも大きい点を始点にする
	for (Map.Entry<Integer, ArrayList<Integer>> entry : map.entrySet()) {
	    if (startKey < entry.getKey()) {
		startKey = entry.getKey();
	    }
	    keyList.add(entry.getKey());
	}
	Collections.sort(keyList, Collections.reverseOrder());
	int i = 0;

	ansList.add(startKey);

	// Y軸の正の向きの単位ベクトル
	double[] prevDir = new double[2];
	prevDir[0] = 0.0;
	prevDir[1] = 1.0;

	while (true) {
	    int fromKey = ansList.get(ansList.size() - 1);
	    int fromX = fromKey / MAX;
	    int fromY = fromKey % MAX;
	    double max = -2;
	    int nextKey = -1;

	    for (int j = i + 1; j < keyList.size(); j++) {
		// 前のベクトルに対して最も内積が大きくなるように次の点を探す
		// そうすると、曲がり角を最大にすることが出来る
		int toKey = keyList.get(j);
		int toX = toKey / MAX;
		int toY = toKey % MAX;

		double[] toDir = direction(fromX, fromY, toX, toY);
		double pro = product(prevDir, toDir);
		if (max < pro) {
		    max = pro;
		    nextKey = toKey;
		    i = j;
		} else if (Math.abs(max - pro) < EPS) {
		    int nx = fromX - nextKey / MAX;
		    int ny = fromY - nextKey % MAX;
		    double ndist = Math.sqrt(nx * nx + ny * ny);
		    int dx = fromX - toX;
		    int dy = fromX - toX;
		    double tdist = Math.sqrt(dx * dx + dy * dy);
		    if (ndist > tdist) {
			max = pro;
			nextKey = toKey;
			i = j;
		    }
		}
	    }

	    if (ansList.contains(nextKey) || nextKey == -1) {
		break;
	    }
	    ansList.add(nextKey);
	    prevDir = direction(fromX, fromY, nextKey / MAX, nextKey % MAX);
	}

	ArrayList<Integer> ans = new ArrayList<>();
	for (Integer key : ansList) {
	    if (key != 0) {
		for (Integer integer : map.get(key)) {
		    if (integer <= N) {
			ans.add(integer);
		    }
		}
	    }
	}
	Collections.sort(ans);
	StringBuilder sb = new StringBuilder();
	for (Integer integer : ans) {
	    sb.append(integer);
	    sb.append(" ");
	}
	System.out.println(sb.toString().trim());

    }

    static double[] direction(int fromX, int fromY, int toX, int toY) {
	// fromからtoへの単位ベクトルを返す
	double dx = toX - fromX;
	double dy = toY - fromY;
	return new double[] { dx / Math.sqrt(dx * dx + dy * dy), dy / Math.sqrt(dx * dx + dy * dy) };
    }

    static double product(double[] a, double[] b) {
	// ベクトル内積を返す
	return a[0] * b[0] + a[1] * b[1];
    }

    static int nextInt() {
	int c;
	try {
	    c = System.in.read();
	    while (c != '-' && (c < '0' || c > '9'))
		c = System.in.read();
	    if (c == '-')
		return -nextInt();
	    int res = 0;
	    while (c >= '0' && c <= '9') {
		res = res * 10 + c - '0';
		c = System.in.read();
	    }
	    return res;
	} catch (IOException e) {
	    // TODO Auto-generated catch block
	    e.printStackTrace();
	}
	return -1;
    }
}