You are given an array consisting of all integers from [l, r] inclusive. For example, if l = 2 and r = 5, the array would be [2, 3, 4, 5]. What's the minimum number of elements you can delete to make the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the array non-zero?

A bitwise AND is a binary operation that takes two equal-length binary representations and performs the AND operation on each pair of the corresponding bits.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t cases follow.

The first line of each test case contains two integers l and r (1 ≤ l ≤ r ≤ 2 ⋅ 10^5) — the description of the array.

Output

For each test case, output a single integer — the answer to the problem.

Example

Input


5
1 2
2 8
4 5
1 5
100000 200000


Output


1
3
0
2
31072

Note

In the first test case, the array is [1, 2]. Currently, the bitwise AND is 0, as 1\ \& \ 2 = 0. However, after deleting 1 (or 2), the array becomes [2] (or [1]), and the bitwise AND becomes 2 (or 1). This can be proven to be the optimal, so the answer is 1.

In the second test case, the array is [2, 3, 4, 5, 6, 7, 8]. Currently, the bitwise AND is 0. However, after deleting 4, 5, and 8, the array becomes [2, 3, 6, 7], and the bitwise AND becomes 2. This can be proven to be the optimal, so the answer is 3. Note that there may be other ways to delete 3 elements.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class B {

    void solve() throws IOException {
        int t = nextInt();
        /*
            EEENE
         */
        int[][] prefixSum = new int[19][200002];
        for (int i = 0; i < 19; ++i) {
            for (int j = 1; j <= 200000; ++j) {
                int bitStatus = (j >> i) & 1;
                bitStatus = bitStatus == 0 ? 1 : 0;
                prefixSum[i][j + 1] = prefixSum[i][j] + bitStatus;
            }
        }
        for (int tt = 0; tt < t; ++tt) {
            int l = nextInt();
            int r = nextInt();
            int minDel = Integer.MAX_VALUE;
            for (int i = 0; i < 19; ++i) {
                int zerosCount = prefixSum[i][r + 1] - prefixSum[i][l];
                if (zerosCount > 0) {
                    minDel = Math.min(minDel, zerosCount);
                } else {
                    minDel = 0;
                }
            }
            writer.println(minDel);
        }
    }

    String nextToken() throws IOException {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(reader.readLine());
        }
        return tokenizer.nextToken();
    }

    int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    StringTokenizer tokenizer;
    PrintWriter writer;
    BufferedReader reader;

    public void run() {
        try {
            writer = new PrintWriter(System.out);
            reader = new BufferedReader(new InputStreamReader(System.in));
            solve();
            reader.close();
            writer.close();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    public static void main(String[] args) {
        new B().run();
    }
}