// Author: btjanaka (Bryon Tjanaka)

// Problem: (UVa) 10026

#include <bits/stdc++.h>

#define FOR(i, a, b) for (int i = a; i < b; ++i)

#define FORe(i, a, b) for (int i = a; i <= b; ++i)

#define PAI(arr, len) /*Print array of integers*/ \

{ \

for (int _i = 0; _i < len; ++_i) { \

if (_i != len - 1) { \

printf("%d ", arr[_i]); \

} else { \

printf("%d", arr[_i]); \

} \

} \

putchar('\n'); \

}

#define PBS(n, len) /*Print a bitset*/ \

{ \

for (int _i = 0; _i < len; ++_i) { \

putchar(n % 2 + '0'); \

n /= 2; \

} \

putchar('\n'); \

}

#define GET(x) scanf("%d", &x)

#define PLN putchar('\n')

#define INF 2147483647

typedef long long ll;

using namespace std;

// Given two jobs, job 1 with fine s1 and days t1, and job 2 with fine s2 and

// days t2, we can compare which one we want to do by looking at s1t2 and s2t1.

// s1t2 is the fine accumulated if we do job 2 first, since we have to pay fine

// s1 for t2 days. s2t1 is the fine accumulated if we do job 1 first, since we

// have to pay fine s2 for t1 days. If s1t2 > s2t1, we will want to do job 1

// first, since that will accumulate the lesser fine (s2t1). We can rearrange

// the equation to be s1/t1 > s2/t2, and we can now use the cost per day ratios

// of each job as a way to compare them with ease. If a job has a smaller s/t

// per day ratio, it should be done last because of the inequality above.

//

// i.e. s1/t1 > s2/t2 --> s1t2 > s2t1 --> do the job with s1 and t1 (job 1)

// first

// first: day per cost ratio

// second: id

pair<float, int> jobs[1010];

int main() {

int ca;

GET(ca);

while (ca--) {

int n;

GET(n);

// Use a days per cost ratio (i.e. inverted) to allow us to use the default

// pair comparator.

FOR(i, 0, n) {

float t, s;

scanf("%f %f", &t, &s);

jobs[i].first = t / s;

jobs[i].second = i + 1;

}

sort(jobs, jobs + n);

// Print results

printf("%d", jobs[0].second);

FOR(i, 1, n) { printf(" %d", jobs[i].second); }

PLN;

if (ca) PLN;

}

return 0;

}