Write SQL query to solve given problem: List the restaurant's ID that has a review greater than the 70% of average review of all American restaurants with street number greater than 2000.. Keep the solution inside sql tag ```sql [SQL-Query] ```
SELECT T1.id_restaurant FROM location AS T1 INNER JOIN generalinfo AS T2 ON T1.id_restaurant = T2.id_restaurant WHERE T2.food_type = 'american' AND T1.street_num > 2000 GROUP BY T1.id_restaurant ORDER BY AVG(T2.review) * 0.7 DESC