Powershell Get Acl:无法验证参数'；路径'；。参数为null或为空

我正在尝试运行下面的代码，但得到错误Get Acl:无法验证参数“Path”上的参数。参数为null或为空。请提供一个不为null或空的参数，然后重试该命令。你知道怎么了吗

$Root = "\\server\share"
$Results = @()
$Folders = Get-ChildItem $Root | Where {$_.PSIsContainer | select fullname
Foreach ($Folder in $Folders) {
    $ACL = Get-Acl $Folder.FullName 
    Foreach ($Group in $ACL.Access) {
        If ("$($Group.IdentityReference)" -like "*-W") {
                $Results += New-Object PSObject -Property @{Folder=$Folder.FullName;Group=$Group.IdentityReference;Members=[string]$(Get-ADGroupMember $($Group.IdentityReference -split "\\")[1] |  Foreach {$_.Name})}
            }
        }
    }
}
$Results | Export-csv c:\pathto\file.csv -NoType

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行中缺少右括号（已修复）：

$Folders = Get-ChildItem $Root | Where {$_.PSIsContainer } | select fullname;