Перенос данных из Excel или таблицы Word в DataGridView простым копированием ячеек

OpenFileDialog dialog = new OpenFileDialog();
dialog.Filter = "MS Excel | *.xlsx";
if (dialog.ShowDialog() == System.Windows.Forms.DialogResult.OK)
{
    OleDbConnection connection = new OleDbConnection(string.Format("Provider=Microsoft.ACE.OLEDB.12.0;Data Source={0};Extended Properties=\"Excel 12.0 Xml;HDR=YES\";", dialog.FileName));
    OleDbDataAdapter adapter = new OleDbDataAdapter("SELECT * FROM [Лист1$]", connection);
    DataTable table = new DataTable();
    adapter.Fill(table);
    dataGridView1.DataSource = table;
}

using Word = Microsoft.Office.Interop.Word;
// ******* 
 
OpenFileDialog dialog = new OpenFileDialog();
dialog.Filter = "MS Word | *.docx";
if (dialog.ShowDialog() == System.Windows.Forms.DialogResult.OK)
{
    Word.Application application = new Word.Application();
    Object FileName = dialog.FileName;
    application.Documents.Open(ref FileName);
    Word.Document document = application.ActiveDocument;
    Word.Table table = document.Tables[1];
    if (table.Rows.Count > 0 && table.Columns.Count > 0)
    {
        List<string> headers = new List<string>();
        DataTable dataTable = new DataTable();
        for (int i = 0; i < table.Columns.Count; i++)
        {
            dataTable.Columns.Add();
            headers.Add(table.Cell(1, i + 1).Range.Text.Trim('\a', '\r', '\n', '\t'));
        }
        for (int i = 0; i < table.Rows.Count - 1; i++)
        {
            string[] row = new string[table.Columns.Count];
            for (int j = 0; j < table.Columns.Count; j++)
                row[j] = table.Cell(i + 2, j + 1).Range.Text.Trim('\a', '\r', '\n', '\t');
            dataTable.Rows.Add(row);
        }
        dataGridView1.DataSource = dataTable;
        for (int i = 0; i < headers.Count; i++)
            dataGridView1.Columns[i].HeaderText = headers[i];
    }
    application.Quit();
}