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## 最终结论
To solve this problem efficiently, we need to count the pairs \((a, b)\) such that \(a + b = a \oplus b\). This condition holds when there is no carry in any bit position during the addition of \(a\) and \(b\). 

The key observation here is that the condition \(a + b = a \oplus b\) is satisfied if and only if there are no positions where both \(a\) and \(b\) have a '1' bit. This means that for each bit position, either \(a\) or \(b\) (or both) should have a '0'.

Given this insight, we can optimize our approach by recognizing that for any two numbers within a certain range \([l, r]\), the maximum number of such pairs occurs when we maximize the number of bits that can be independently set between the two numbers. This happens when we consider the largest power of 2 that fits within the range from \(l\) to \(r\).

Here's how you can implement this:

```python
def count_pairs(l, r):
    # Find the maximum x such that 2^x - 1 fits in [l, r]
    max_x = 0
    while (1 << max_x) <= r:
        max_x += 1
    
    # Calculate the result based on the highest power of 2 less than or equal to r
    # The number of valid pairs is determined by how many bits can be independently set
    # between l and r.
    return (1 << (max_x - 1)) * ((1 << (max_x - 1)) - 1)

t = int(input())
results = []
for _ in range(t):
    l, r = map(int, input().split())
    results.append(count_pairs(l, r))

for result in results:
    print(result)
```

This code efficiently calculates how many pairs satisfy the condition by leveraging bit manipulation and understanding how XOR works with respect to binary representations. The complexity is reduced significantly by avoiding unnecessary pair checks and focusing on bit-level operations.