For problem statement at 1000-1999/1900-1999/1970-1979/1970/problemA2.txt this is a correct solution, but verifier at 1000-1999/1900-1999/1970-1979/1970/verifierA2.go ends with case 2 failed: runtime error: exit status 2
panic: runtime error: index out of range [1] with length 1

goroutine 1 [running]:
main.main()
	/tmp/build-1838704784/solution.go:149 +0x698

exit status 1 can you fix the verifier? ```go
package main

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

var (
	s     string
	n     int
	memo  map[[3]int]bool
	cuts  map[[3]int]int
)

// solve determines if the suffix s[idx:] can be partitioned starting with a block 
// that satisfies the required number of closing parentheses reqClosed, given that
// the previous block contributed openPrev open parentheses.
// returns true if a valid partition exists.
func solve(idx, reqClosed, openPrev int) bool {
	// Base case: if we used all characters
	if idx == n {
		// Valid only if no more closed parentheses are required
		// and no open parentheses from previous level remain unhandled (though openPrev=0 is implied by reqClosed logic effectively)
		return reqClosed == 0 && openPrev == 0
	}
	
	state := [3]int{idx, reqClosed, openPrev}
	if v, ok := memo[state]; ok {
		return v
	}

	currentClosed := 0
	currentOpen := 0
	
	res := false
	
	// Iterate through possible end positions for the current block
	// The current block MUST contain exactly reqClosed ')' characters.
	// We can include any number of '(' characters.
	// Once we exceed reqClosed ')', the block is invalid.
	for i := idx; i < n; i++ {
		if s[i] == ')' {
			currentClosed++
		} else {
			currentOpen++
		}
		
		if currentClosed > reqClosed {
			break
		}
		
		if currentClosed == reqClosed {
			// Found a candidate block s[idx...i]
			// Length of current block
			cCurr := (i - idx + 1)
			
			// Calculate requirement for the next block (level + 1)
			// Formula: c_curr = open_prev + closed_next
			// So closed_next = c_curr - open_prev
			reqNext := cCurr - openPrev
			
			if reqNext >= 0 {
				// Try to solve for the next level
				if solve(i+1, reqNext, currentOpen) {
					cuts[state] = i + 1
					res = true
					break // Found a valid path, break since solution is unique
				}
			}
		}
	}
	
	memo[state] = res
	return res
}

func main() {
	// 1. Read Input
	reader := bufio.NewReader(os.Stdin)
	line, _ := reader.ReadString('\n')
	s = ""
	for _, c := range line {
		if c == '(' || c == ')' {
			s += string(c)
		}
	}
	n = len(s)
	
	// 2. Initialize DP/Memoization
	memo = make(map[[3]int]bool)
	cuts = make(map[[3]int]int)
	
	// 3. Find the partition of s into blocks corresponding to levels
	// Level 0 starts with reqClosed = 0 because level 0 contains items with balance 0.
	// Since prefix balance >= 0, items with balance 0 must be '(', so they cannot be ')'.
	// Thus reqClosed for level 0 is 0. openPrev is 0.
	if !solve(0, 0, 0) {
		// This should not happen given the problem guarantees
		return
	}
	
	// 4. Reconstruct the blocks
	var blocks []string
	currIdx := 0
	currReq := 0
	currOpenPrev := 0
	
	for currIdx < n {
		state := [3]int{currIdx, currReq, currOpenPrev}
		nextIdx := cuts[state]
		
		block := s[currIdx:nextIdx]
		blocks = append(blocks, block)
		
		opCurr := 0
		for _, c := range block {
			if c == '(' {
				opCurr++
			}
		}
		
		currReq = len(block) - currOpenPrev
		currOpenPrev = opCurr
		currIdx = nextIdx
	}
	
	// 5. Process blocks to get items for each level
	// The problem states that within a level, items are sorted by original position descending.
	// To reconstruct t (positions 1, 2, ...), we need items with smallest original positions first.
	// So we reverse the characters in each block.
	items := make([][]byte, len(blocks))
	for i, b := range blocks {
		chars := []byte(b)
		for left, right := 0, len(chars)-1; left < right; left, right = left+1, right-1 {
			chars[left], chars[right] = chars[right], chars[left]
		}
		items[i] = chars
	}
	
	// 6. Simulate the walk to reconstruct t
	// We start at level 0. For each step, we take the next available character from current level.
	// If char is '(', we go to level + 1. If ')', we go to level - 1.
	t := make([]byte, 0, n)
	lvl := 0
	ptr := make([]int, len(blocks))
	
	for i := 0; i < n; i++ {
		char := items[lvl][ptr[lvl]]
		ptr[lvl]++
		t = append(t, char)
		
		if char == '(' {
			lvl++
		} else {
			lvl--
		}
	}
	
	fmt.Println(string(t))
}
```