underscore: fix chain with Dictionary<T>

I added a test for this, it would previously fail to compile.

Also, using `_.values()` on a Dictionary would lose strong typing on the result, returning `any[]`. This now properly returns `T[]` in that case.

underscore/underscore-tests.ts

@@ -479,3 +479,20 @@ _.chain(obj).map(function (value, key) {
     empty[key] = value;
     console.log("vk", value, key);
 });
+
+function strong_typed_values_tests() {
+  var dictionaryLike: { [k: string] : {title: string, value: number} } = {
+         'test' : { title: 'item1', value: 5 },
+         'another' : { title: 'item2', value: 8 },
+         'third' : { title: 'item3', value: 10 }
+      },
+      empty = {};
+
+  _.chain(dictionaryLike).values().filter((r) => {
+    return r.value >= 8;
+  }).map((r) => {
+    return [r.title, true];
+  }).object().value();
+
+  _.values<{title: string, value: number}>(dictionaryLike);
+}

underscore/underscore.d.ts

@@ -1192,6 +1192,13 @@ interface UnderscoreStatic {
 	**/
 	keys(object: any): string[];
 
+	/**
+	* Return all of the values of the object's properties.
+	* @param object Retrieve the values of all the properties on this object.
+	* @return List of all the values on `object`.
+	**/
+	values<T>(object: _.Dictionary<T>): T[];
+
 	/**
 	* Return all of the values of the object's properties.
 	* @param object Retrieve the values of all the properties on this object.
@@ -1641,6 +1648,7 @@ interface UnderscoreStatic {
 	* @return Wrapped `obj`.
 	**/
 	chain<T>(obj: T[]): _Chain<T>;
+	chain<T>(obj: _.Dictionary<T>): _Chain<T>;
 	chain<T extends {}>(obj: T): _Chain<T>;
 }