Problem: Recently Petya walked in the forest and found a magic stick.

Since Petya really likes numbers, the first thing he learned was spells for changing numbers. So far, he knows only two spells that can be applied to a positive integer: 

  1. If the chosen number a is even, then the spell will turn it into 3a/2; 
  2. If the chosen number a is greater than one, then the spell will turn it into a-1. 



Note that if the number is even and greater than one, then Petya can choose which spell to apply.

Petya now has only one number x. He wants to know if his favorite number y can be obtained from x using the spells he knows. The spells can be used any number of times in any order. It is not required to use spells, Petya can leave x as it is.

Input

The first line contains single integer T (1 ≤ T ≤ 10^4) — the number of test cases. Each test case consists of two lines.

The first line of each test case contains two integers x and y (1 ≤ x, y ≤ 10^9) — the current number and the number that Petya wants to get.

Output

For the i-th test case print the answer on it — YES if Petya can get the number y from the number x using known spells, and NO otherwise.

You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).

Example

Input


7
2 3
1 1
3 6
6 8
1 2
4 1
31235 6578234


Output


YES
YES
NO
YES
NO
YES
YES

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int ed, ex, ey;
void extendedEuclid(long long int A, long long int B) {
  if (B == 0) {
    ed = A;
    ex = 1;
    ey = 0;
  } else {
    extendedEuclid(B, A % B);
    long long int temp = ex;
    ex = ey;
    ey = temp - (A / B) * ey;
  }
}
long long int modInverse(long long int A, long long int M) {
  extendedEuclid(A, M);
  return (ex % M + M) % M;
}
long long int todec(long long int b, string s) {
  long long int ans = 0, c = 1, l = s.length();
  reverse(s.begin(), s.end());
  for (long long int i = 0; i < l; i++) {
    if (s[i] <= '9')
      ans += c * (s[i] - '0');
    else
      ans += c * (s[i] - 'A' + 10);
    c = c * b;
  }
  return ans;
}
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x = x % 1000000007;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res;
}
void binary_rep(long long int n) {
  long long int a = 1;
  while (a * 2 <= n) a = a * 2;
  while (a) {
    if (a & n)
      cout << 1;
    else
      cout << 0;
    a /= 2;
  }
}
long long int nextPerfectSquare(long long int N) {
  long long int nextN = floor(sqrt(N)) + 1;
  return nextN * nextN;
}
long double dist(long long int x1, long long int y1, long long int x2,
                 long long int y2) {
  long long int a = (x1 - x2) * (x1 - x2);
  long long int b = (y1 - y2) * (y1 - y2);
  return sqrtl(a + b);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long int x, y;
    cin >> x >> y;
    if (x >= y)
      cout << "YES";
    else {
      if (x == 1)
        cout << "NO";
      else {
        if (x == 2 && y > 3)
          cout << "NO";
        else if (x == 3 && y > 3)
          cout << "NO";
        else
          cout << "YES";
      }
    }
    cout << "\n";
  }
  return 0;
}