How to detect button press for 5 seconds and then open a dialog window

using System;
using System.Windows.Forms;

namespace OpenFormAfter5Seconds
{
    static class Program
    {
        [STAThread]
        static void Main()
        {
            // Create a test form with a button
            Form form = new Form();
            Button button = new Button();
            button.Text = "Press for 5 s";
            button.Dock = DockStyle.Fill;
            form.Controls.Add(button);

            // Create a timer for meassuring the duration
            Timer timer = new Timer();
            timer.Interval = 5000; // 5 Seconds

            // Process Button MouseDown 
            button.MouseDown += delegate(object sender, MouseEventArgs mea)
            {
                // Start the timer 
                timer.Start();
            };

            // Process Button MouseUp 
            button.MouseUp += delegate(object sender, MouseEventArgs mea)
            {
                // Stop the timer 
                timer.Stop();
            };

            // If Timer.Tick fires (after timer intervall has elapsed)
            timer.Tick += delegate(object sender, EventArgs e)
            {
                // ... stop the timer - it shouldn't fire again
                timer.Stop();

                // ... Open the new dialog - let's use a MessageBox here for demonstration
                MessageBox.Show(form, "You Pressed the Button for 5 Seconds!", "Dialog");
            };

            // run the example
            Application.Run(form);
        }
    }
}