+ for(i = 1, nn = n; i < n; i++)
+ if(info[i - 1].x_org == info[i].x_org && info[i - 1].y_org == info[i].y_org
+ && info[i - 1].width == info[i].width && info[i - 1].height == info[i].height)
+ --nn;
+ n = nn; /* we only consider unique geometries as separate screens */