For problem statement at 1000-1999/1400-1499/1410-1419/1419/problemA.txt this is a correct solution, but verifier at 1000-1999/1400-1499/1410-1419/1419/verifierA.go ends with All tests passed can you fix the verifier? ```go
package main

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

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

	var t int
	fmt.Fscan(reader, &t)

	for ; t > 0; t-- {
		var n int
		fmt.Fscan(reader, &n)
		var s string
		fmt.Fscan(reader, &s)

		// Raze picks from odd positions (1-indexed), i.e., indices 0, 2, 4, ... (0-indexed)
		// Breach picks from even positions (1-indexed), i.e., indices 1, 3, 5, ... (0-indexed)
		// Raze wants the last remaining digit to be odd
		// Breach wants the last remaining digit to be even

		// Count digits on odd positions (1-indexed) and even positions (1-indexed)
		// Odd positions (1-indexed) = indices 0, 2, 4, ... (0-indexed)
		// Even positions (1-indexed) = indices 1, 3, 5, ... (0-indexed)

		oddPosCount := 0  // count of digits at odd positions (1-indexed)
		evenPosCount := 0 // count of digits at even positions (1-indexed)

		for i := 0; i < n; i++ {
			if (i+1)%2 == 1 {
				oddPosCount++
			} else {
				evenPosCount++
			}
		}

		// Raze removes from odd positions, Breach removes from even positions
		// They alternate, Raze first. Total turns = n - 1
		// Raze makes ceil((n-1)/2) turns, Breach makes floor((n-1)/2) turns

		// Raze removes from odd positions: she removes (oddPosCount - x) digits from odd positions
		// where x is how many odd-position digits remain
		// Breach removes from even positions: he removes (evenPosCount - y) digits from even positions

		// After all turns, exactly 1 digit remains.
		// Raze makes ceil((n-1)/2) removals from odd positions
		// Breach makes floor((n-1)/2) removals from even positions

		// But wait: Raze can only remove from odd positions. She has oddPosCount digits available.
		// She needs to make ceil((n-1)/2) removals. But she might not have enough odd-position digits.
		// Actually the problem says "It can be proved that before the end each agent has ability to make a turn"
		// So we can trust counts work out.

		// Raze removes ceil((n-1)/2) digits from odd positions
		// Breach removes floor((n-1)/2) digits from even positions

		// After removal:
		// Remaining odd-position digits: oddPosCount - ceil((n-1)/2)
		// Remaining even-position digits: evenPosCount - floor((n-1)/2)
		// Total remaining = 1

		razeTurns := (n - 1 + 1) / 2 // ceil((n-1)/2)
		breachTurns := (n - 1) / 2    // floor((n-1)/2)

		remainOdd := oddPosCount - razeTurns
		remainEven := evenPosCount - breachTurns

		// Exactly one of remainOdd, remainEven should be 1 and the other 0
		_ = remainEven

		if remainOdd == 1 {
			// The last digit is from odd positions. Raze wants it odd.
			// Raze is removing digits from odd positions to leave one she likes (odd digit)
			// She plays optimally: she removes even digits from odd positions first.
			// She wants to keep an odd digit. Does any odd-position digit have an odd value?
			hasOdd := false
			for i := 0; i < n; i += 2 {
				if (s[i]-'0')%2 == 1 {
					hasOdd = true
					break
				}
			}
			if hasOdd {
				fmt.Fprintln(writer, 1)
			} else {
				fmt.Fprintln(writer, 2)
			}
		} else {
			// The last digit is from even positions. Breach controls it.
			// Breach wants to keep an even digit.
			hasEven := false
			for i := 1; i < n; i += 2 {
				if (s[i]-'0')%2 == 0 {
					hasEven = true
					break
				}
			}
			if hasEven {
				fmt.Fprintln(writer, 2)
			} else {
				fmt.Fprintln(writer, 1)
			}
		}
	}
}
```