Looking at this problem, I need to:
1. Build array c[1] = [1, a_2, a_3, ..., a_n] since m=1
2. Find minimum operations to make all remaining pairs satisfy a[i] < b[i]
3. This is equivalent to maximizing the number of valid pairs we can form

The approach:
- Sort both arrays
- Use a greedy two-pointer technique: for each smallest remaining a element, find the smallest b element that's larger than it
- This maximizes pairs since using smaller b values when possible leaves larger b values for larger a elements

Let me trace through the example:
- n=2, a_2=1, b=[3,2]
- c[1] = [1, 1], after sorting: [1, 1]
- b sorted: [2, 3]
- Pointer on a: 1, Pointer on b: 2 → 1<2 ✓, pair=1
- Pointer on a: 1, Pointer on b: 3 → 1<3 ✓, pair=2
- pairs=2, operations = 2-2 = 0 ✓

```go
package main

import (
    "bufio"
    "fmt"
    "os"
    "sort"
)

func main() {
    reader := bufio.NewReader(os.Stdin)
    writer := bufio.NewWriter(os.Stdout)
    defer writer.Flush()
    
    var t int
    fmt.Fscan(reader, &t)
    
    for tc := 0; tc < t; tc++ {
        var n, m int
        fmt.Fscan(reader, &n, &m)
        
        a := make([]int, n)
        a[0] = 1
        for i := 1; i < n; i++ {
            fmt.Fscan(reader, &a[i])
        }
        
        b := make([]int, n)
        for i := 0; i < n; i++ {
            fmt.Fscan(reader, &b[i])
        }
        
        sort.Ints(a)
        sort.Ints(b)
        
        pa, pb := 0, 0
        pairs := 0
        for pa < n && pb < n {
            if a[pa] < b[pb] {
                pairs++
                pa++
                pb++
            } else {
                pb++
            }
        }
        
        fmt.Fprintln(writer, n-pairs)
    }
}
```