Problem: In 2N - 1 boxes there are apples and oranges. Your task is to choose N boxes so, that they will contain not less than half of all the apples and not less than half of all the oranges.

Input

The first input line contains one number T — amount of tests. The description of each test starts with a natural number N — amount of boxes. Each of the following 2N - 1 lines contains numbers ai and oi — amount of apples and oranges in the i-th box (0 ≤ ai, oi ≤ 109). The sum of N in all the tests in the input doesn't exceed 105. All the input numbers are integer.

Output

For each test output two lines. In the first line output YES, if it's possible to choose N boxes, or NO otherwise. If the answer is positive output in the second line N numbers — indexes of the chosen boxes. Boxes are numbered from 1 in the input order. Otherwise leave the second line empty. Separate the numbers with one space.

Examples

Input

2
2
10 15
5 7
20 18
1
0 0


Output

YES
1 3
YES
1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:100000000")
using namespace std;
const long long mod = 1000000007;
const long long md = 1234567891;
const long long bil = 1000000000;
const char aa = 'a';
const double eps = 1E-9;
const int mil = 1000000;
const int ths = 1000;
const int maxn = 1001;
const int dr[] = {0, -1, 0, 1};
const int dc[] = {1, 0, -1, 0};
const int mxx = 32;
const int inf = (int)1E9;
template <typename T>
void printV(vector<T>& v) {
  for (int i = 0; i < (int)(v.size()); i++) cout << v[i] << " ";
  cout << endl;
}
template <typename T>
void printVs(vector<vector<T> >& v) {
  for (int i = 0; i < (int)(v.size()); i++) printV(v[i]);
}
long long gcd(long long m, long long n) {
  if (m == 0 || n == 0) return m + n;
  return gcd(n, m % n);
}
const int mxn = 100001;
vector<int> getDigit(long long N, int bs) {
  vector<int> res;
  while (N) {
    res.push_back(N % bs);
    N /= bs;
  }
  {
    for (int i = 0; i < (int)(res.size() / 2); i++)
      swap(res[i], res[res.size() - i - 1]);
  };
  return res;
}
int main() {
  int T;
  cin >> T;
  vector<vector<int> > res;
  for (int t = 0; t < (int)(T); t++) {
    int N;
    vector<vector<int> > A;
    cin >> N;
    for (int i = 0; i < (int)(2 * N - 1); i++) {
      int a, b;
      cin >> a >> b;
      vector<int> v;
      v.push_back(a);
      v.push_back(b);
      v.push_back(i + 1);
      A.push_back(v);
    }
    sort(A.begin(), A.end());
    long long sumOdd = 0, sumEven = 0;
    for (int i = 0; i < (int)(A.size() - 1); i++) {
      if (i % 2)
        sumOdd += A[i][1];
      else
        sumEven += A[i][1];
    }
    vector<int> v;
    if (sumOdd > sumEven) {
      for (int i = (1); i < (int)(A.size() - 1); i++)
        if (i % 2) v.push_back(A[i][2]);
    } else
      for (int i = 0; i < (int)(A.size() - 1); i++)
        if (i % 2 == 0) v.push_back(A[i][2]);
    v.push_back(A[2 * N - 2][2]);
    assert(v.size() == N);
    res.push_back(v);
  }
  for (int t = 0; t < (int)(T); t++) {
    cout << "YES" << endl;
    printV(res[t]);
  }
}