Given a positive integer n, there exists a 0-indexed array called powers, composed of the minimum number of powers of 2 that sum to n. The array is sorted in non-decreasing order, and there is only one way to form the array.

You are also given a 0-indexed 2D integer array queries, where queries[i] = [lefti, righti]. Each queries[i] represents a query where you have to find the product of all powers[j] with lefti <= j <= righti.

Return an arrayanswers, equal in length toqueries, whereanswers[i]is the answer to theithquery. Since the answer to the ith query may be too large, each answers[i] should be returned modulo109 + 7.

Input: n = 15, queries = [[0,1],[2,2],[0,3]] Output: [2,4,64] Explanation: For n = 15, powers = [1,2,4,8]. It can be shown that powers cannot be a smaller size. Answer to 1st query: powers[0] * powers[1] = 1 * 2 = 2. Answer to 2nd query: powers[2] = 4. Answer to 3rd query: powers[0] * powers[1] * powers[2] * powers[3] = 1 * 2 * 4 * 8 = 64. Each answer modulo 109 + 7 yields the same answer, so [2,4,64] is returned. 

Input: n = 2, queries = [[0,0]] Output: [2] Explanation: For n = 2, powers = [2]. The answer to the only query is powers[0] = 2. The answer modulo 109 + 7 is the same, so [2] is returned. 

• 1 <= n <= 109
• 1 <= queries.length <= 105
• 0 <= starti <= endi < powers.length

implSolution{pubfnproduct_queries(n:i32,queries:Vec<Vec<i32>>) -> Vec<i32>{letmut powers = vec![];letmut answers = vec![1_i64; queries.len()];for i in0..30{if n &(1 << i) != 0{ powers.push(1 << i);}}for i in0..queries.len(){for j in queries[i][0]asusize..=queries[i][1]asusize{ answers[i] = (answers[i]* powers[j]) % 1_000_000_007;}} answers.into_iter().map(|x| x asi32).collect()}}