For problem statement at 1000-1999/1100-1199/1170-1179/1172/problemA.txt this is a correct solution, but verifier at 1000-1999/1100-1199/1170-1179/1172/verifierA.go ends with All tests passed can you fix the verifier? ```go
package main

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

func main() {
	scanner := bufio.NewScanner(os.Stdin)
	scanner.Split(bufio.ScanWords)

	if !scanner.Scan() {
		return
	}
	n, _ := strconv.Atoi(scanner.Text())

	// Read cards in hand (we don't strictly need their values for the logic, 
	// knowing who is in the pile is sufficient as the rest are in hand)
	for i := 0; i < n; i++ {
		scanner.Scan()
	}

	b := make([]int, n)
	pos := make([]int, n+1)
	// Initialize pos with -1 (representing "in hand")
	for i := 0; i <= n; i++ {
		pos[i] = -1
	}

	for i := 0; i < n; i++ {
		scanner.Scan()
		val, _ := strconv.Atoi(scanner.Text())
		b[i] = val
		if val != 0 {
			pos[val] = i
		}
	}

	// Strategy 1: Check if the pile already contains a suffix 1, 2, ..., k at the bottom.
	// If so, we might just need to append k+1, ..., n.
	idx1 := pos[1]
	if idx1 != -1 {
		k := n - idx1
		// Check if b[idx1...n-1] is 1, 2, ..., k
		isSuffix := true
		for i := 0; i < k; i++ {
			if b[idx1+i] != 1+i {
				isSuffix = false
				break
			}
		}

		if isSuffix {
			// Check if we can play k+1, ..., n immediately
			// Card val needs to be played at operation (val - k).
			// If card val is in the pile at index p, it becomes available at operation p + 2.
			// Requirement: p + 2 <= val - k  =>  p - val + k + 2 <= 0
			possible := true
			for val := k + 1; val <= n; val++ {
				p := pos[val]
				if p != -1 {
					if p-val+k+2 > 0 {
						possible = false
						break
					}
				}
			}
			if possible {
				fmt.Println(n - k)
				return
			}
		}
	}

	// Strategy 2: We cannot extend an existing suffix starting with 1.
	// We must extract cards until we can start playing 1, 2, ..., n.
	// Suppose we play card 1 at operation S+1. Then card x is played at S+x.
	// Card x at pile index p is available at p+2.
	// Requirement: p + 2 <= S + x  =>  S >= p - x + 2
	maxWait := 0
	for i := 0; i < n; i++ {
		if b[i] != 0 {
			val := b[i]
			req := i - val + 2
			if req > maxWait {
				maxWait = req
			}
		}
	}

	fmt.Println(n + maxWait)
}
```