Открыть текстовый файл для чтения из программы (без окна OpenFileDialog) - C#

Stream myStream3;
            OpenFileDialog openFileDialog1 = new OpenFileDialog();
 
            openFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
            openFileDialog1.FilterIndex = 2;
            openFileDialog1.RestoreDirectory = true;
 
            if (openFileDialog1.ShowDialog() == DialogResult.OK)
            {
                if ((myStream3 = openFileDialog1.OpenFile()) != null)
                {
                    StreamReader str = new StreamReader(myStream3);
                    for (int i = 0; i < 10; i++)
                    {
                        for (int j = 0; j < 2; j++)
                        {
                            
                            kouc44[i][j] = Convert.ToInt16(str.ReadLine());
                                }
                       
                    }
str.Close();
                    
                    myStream3.Close();
                }
            }

StreamReader  str= new StreamReader(@"C:\...\...\bin\Debug\данные.txt");
                    for (int i = 0; i < 10; i++)
                    {
                        for (int j = 0; j < 2; j++)
                        {
                            
                            kouc44[i][j] = Convert.ToInt16(str.ReadLine());
                                }
                          }
 str.Close();