Error trying to deserialize an XML file with openfiledialog.filename?

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C#

XML

WinForms

Hello, I have an error in my code, i want to deserialize different XML Files and open it in textboxes, the problem i had is when it takes the file path, i Think that there is where the error is, because it takes a one exact file path(ex.):

FileStream FS = new FileStream("example.analy", FileMode.Open, FileAccess.Read);

But i replaced it with this

FileStream FS = new FileStream(openFileDialog1.FileName, FileMode.Open, FileAccess.Read);

To read the files that were selected in the openfiledialog.
This is my complete code:

if (openFileDialog1.ShowDialog() == DialogResult.OK)
          {
              if (File.Exists(openFileDialog1.FileName))
              {
              if ((openFileDialog1.OpenFile()) != null)
              {
                      if (openFileDialog1.FileName != null)
                      {
                          openFileDialog1.Filter = "Analytica Files | *.analy";
                          XmlSerializer XS = new XmlSerializer(typeof(Reminders.Information));

                          FileStream FS = new FileStream(openFileDialog1.FileName, FileMode.Open, FileAccess.Read);
                          Information Info = (Information)XS.Deserialize(FS);
                          XS.Deserialize(FS);
                          Settings.Default.Counter++;
                          Settings.Default.Save();

The problem is that when I try to open a file, appears the next error:
System.InvalidOperationException: 'Error en el documento XML (0, 0).'
Inner Exception
XmlException: Missing Root Element.

What I have tried:

This is my complete code:
<pre>  if (openFileDialog1.ShowDialog() == DialogResult.OK)
            {
                if (File.Exists(openFileDialog1.FileName))
                {              
                if ((openFileDialog1.OpenFile()) != null)
                {
                        if (openFileDialog1.FileName != null)
                        {
                            openFileDialog1.Filter = "Analytica Files | *.analy";
                            XmlSerializer XS = new XmlSerializer(typeof(Reminders.Information));
                         
                            FileStream FS = new FileStream(openFileDialog1.FileName, FileMode.Open, FileAccess.Read);
                            Information Info = (Information)XS.Deserialize(FS);
                            XS.Deserialize(FS);
                            Settings.Default.Counter++;
                            Settings.Default.Save();