Python: Fix cryptography modeling

The old code was my own suggestion, that I thought would just work, but
was also slightly skeptical about.

I tested out whether it works with the code below

```codeql
predicate foo(int input, string res) {
  input = 1 and res = "that was one"
}

from int input, string res
where
  input in [1, 2] and
  if foo(input, res)
  then any()
  else res = "not one"
select input, res
```

which gave the 3 results

```
1 |	that was one
1 |	not one
2 |	not one
```

only by rewriting the code to be the one below, did I get down to the 2
results I actually wanted. So I've done the same kind of rewrite in the
commit.

```codeql
predicate foo(int input, string res) {
  input = 1 and res = "that was one"
}

from int input, string res
where
  input in [1, 2] and
  if foo(input, _)
  then foo(input, res)
  else res = "not one"
select input, res
```

Author: RasmusWL

python/ql/lib/semmle/python/frameworks/Cryptography.qll

@@ -195,8 +195,8 @@ private module CryptographyModel {
             call.getArg(0), call.getArgByName("algorithm")
           ] and
         exists(DataFlow::Node modeArg | modeArg in [call.getArg(1), call.getArgByName("mode")] |
-          if modeArg = modeClassRef(modeName).getReturn().getAUse()
-          then any()
+          if modeArg = modeClassRef(_).getReturn().getAUse()
+          then modeArg = modeClassRef(modeName).getReturn().getAUse()
           else modeName = "<None or unknown>"
         )
       )