Function argument which transforms in ST Monad

Question

How can I write the following function tt, which has currently type error:

t :: Int
t = runST $ do
  ref <- newSTRef 10
  readSTRef ref

tt :: (STRef s a -> ST s a) -> Int
tt f = runST $ do
  ref <- newSTRef 10
  f ref

ttTest = tt readSTRef

I thought that inside runST in tt, state variable s can be threaded into function f, but following compiler error tells me that i'm wrong:

transform.hs:50:3: Couldn't match type `s' with `s1' …
  `s' is a rigid type variable bound by
      the type signature for tt :: (STRef s a -> ST s a) -> Int
      at transform.hs:47:7
  `s1' is a rigid type variable bound by
       a type expected by the context: ST s1 Int
       at transform.hs:48:8
Expected type: ST s1 Int
  Actual type: ST s a
Relevant bindings include
  ref :: STRef s1 a
    (bound at transform.hs:49:3)
  f :: STRef s a -> ST s a
    (bound at transform.hs:48:4)
  tt :: (STRef s a -> ST s a) -> Int
    (bound at transform.hs:48:1)
In a stmt of a 'do' block: f ref
In the second argument of `($)', namely
  `do { ref <- newSTRef 10;
        f ref }'
transform.hs:50:3: Couldn't match type `a' with `Int' …
  `a' is a rigid type variable bound by
      the type signature for tt :: (STRef s a -> ST s a) -> Int
      at transform.hs:47:7
Expected type: ST s1 Int
  Actual type: ST s a
Relevant bindings include
  ref :: STRef s1 a
    (bound at transform.hs:49:3)
  f :: STRef s a -> ST s a
    (bound at transform.hs:48:4)
  tt :: (STRef s a -> ST s a) -> Int
    (bound at transform.hs:48:1)
In a stmt of a 'do' block: f ref
In the second argument of `($)', namely
  `do { ref <- newSTRef 10;
        f ref }'

Any comments will be appreciated deeply.

Answer 1

The answer was posted pseudo-anonymously:

tt :: (forall s . STRef s Int -> ST s Int) -> Int.