title: SWRevealViewController panGestureRecognizer with UIScreenEdgeGestureRecognizer
body: <p>I've been looking for solution for this question,but didnt find any one. So I use SWRevealViewController for sidebar menu in my project but I want that side menu would open when user beging to touch from left edge of the view. I want this because I have another swipeGestureRecognizer for left and right swipe.For adding left and right swipes I'm use this code:</p>
  
  <pre><code>    UISwipeGestureRecognizer *swipeRight = [[UISwipeGestureRecognizer alloc]initWithTarget:self action:@selector(swipedToRight:)];
          swipeRight.direction = UISwipeGestureRecognizerDirectionRight;
  
     UISwipeGestureRecognizer *swipeLeft = [[UISwipeGestureRecognizer alloc] initWithTarget: self action: @selector(swipedToLeft:)];
         swipeLeft.direction = UISwipeGestureRecognizerDirectionLeft;
  </code></pre>
  
  <p>So my question is how can I make that side menu would open only when touches begins from left edge?</p>
  
  <p><strong>UPD 1</strong></p>
  
  <p>I use this code to open side menu by panning at any part of the screen</p>
  
  <pre><code>[self.view addGestureRecognizer: self.revealViewController.panGestureRecognizer];
  </code></pre>
  
  <p>And I have recognizer to swipe to right. The problem is once it open menu and once executes swipe's action. I want to allow to open menu only when touches is from left edge</p>
  
  <p><strong>UPD 2 Solved</strong></p>
  
  <p>I solved it by setting draggable border width:</p>
  
  <pre><code>self.revealViewController.draggableBorderWidth = self.view.frame.size.width / 2;
  </code></pre>
  
  <p>And now gesture from edge to center opens menu, and from center to right border executes swipe's action</p>