```go
package main

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

func main() {
	in := bufio.NewReader(os.Stdin)
	out := bufio.NewWriter(os.Stdout)
	defer out.Flush()

	var t int
	if _, err := fmt.Fscan(in, &t); err != nil {
		return
	}

	for k := 0; k < t; k++ {
		var n int
		if _, err := fmt.Fscan(in, &n); err != nil {
			break
		}
		var c string
		if _, err := fmt.Fscan(in, &c); err != nil {
			break
		}

		// If n is odd, the answer is always YES
		if n%2 != 0 {
			fmt.Fprintln(out, "YES")
			continue
		}

		// If n is even, check for obstructions
		hasRR := false
		hasBB := false
		countR := 0
		countB := 0

		for i := 0; i < n; i++ {
			if c[i] == 'R' {
				countR++
			} else {
				countB++
			}

			// Check adjacent edge colors
			next := i + 1
			if next == n {
				next = 0
			}

			if c[i] == c[next] {
				if c[i] == 'R' {
					hasRR = true
				} else {
					hasBB = true
				}
			}
		}

		if hasRR && hasBB {
			// Contains both "RR" and "BB" -> disconnected
			fmt.Fprintln(out, "NO")
		} else if !hasRR && !hasBB {
			// Alternating colors -> disconnected
			fmt.Fprintln(out, "NO")
		} else {
			// Only "RR" or only "BB" exists.
			// The isolated characters must appear an odd number of times (or 0) for YES.
			if hasRR {
				// Only RR exists, check count of B
				// If countB is even and > 0, it's NO
				if countB > 0 && countB%2 == 0 {
					fmt.Fprintln(out, "NO")
				} else {
					fmt.Fprintln(out, "YES")
				}
			} else {
				// Only BB exists, check count of R
				if countR > 0 && countR%2 == 0 {
					fmt.Fprintln(out, "NO")
				} else {
					fmt.Fprintln(out, "YES")
				}
			}
		}
	}
}
```