Support 'is not _' expression syntax

[tjsmith-meta]

Here's the example template that works in Python jinja2 -- 'is not defined'.

{% if foo is not defined %}
not defined
{% endif %}
{% set foo = False %}
{% if foo is not defined %}
  {% set foo = True %}
{% endif %}
{{ foo }}

I realize there is an 'undefined' operator which is already supported, but 'not defined' appears to be a valid formulation as well.

Here's the change I have for reference that seems to do the trick for me.

diff --git a/jinja2cpp/src/expression_parser.cpp b/jinja2cpp/src/expression_parser.cpp
--- a/jinja2cpp/src/expression_parser.cpp
+++ b/jinja2cpp/src/expression_parser.cpp
@@ -151,7 +151,14 @@
             break;
         case Keyword::Is:
         {
+            bool wrapWithNot = false;
+
             Token nextTok = lexer.NextToken();
+            if (lexer.GetAsKeyword(nextTok) == Keyword::LogicalNot) {
+              wrapWithNot = true;
+              nextTok = lexer.NextToken();
+            }
+
             if (nextTok != Token::Identifier)
                 return MakeParseError(ErrorCode::ExpectedIdentifier, nextTok);

@@ -164,7 +171,12 @@
             if (!params)
                 return params.get_unexpected();

-            return std::make_shared<IsExpression>(*left, std::move(name), std::move(*params));
+            auto isExpr = std::make_shared<IsExpression>(*left, std::move(name), std::move(*params));
+            if (wrapWithNot) {
+              return std::make_shared<UnaryExpression>(UnaryExpression::Operation::LogicalNot, isExpr);
+            } else {
+              return isExpr;
+            }
         }
         default:
             lexer.ReturnToken();