public static void solve() {
        int n = scanner.nextInt();
        int[] a = new int[n];
        
        // the problem says I get to multiply any number I want by 2 as long
        // as I divide another even number by two, so I'll count how many times
        // I can do this (how many multiplications by 2 I get to have) and store it in a variable
        // To do this, I keep peeling off all the twos from each a[i] as long as it remains even
        int twos = 0;
        for (int i = 0; i < n; i++) {
            a[i] = scanner.nextInt();
            while (a[i] >= 2 && a[i] % 2 == 0) {
                twos++;
                a[i] /= 2;
            }
        }
        // sorting the array helps with the next step
        Arrays.sort(a);
        int ans = 1;
        // I binary search in the range from 1 to n to find the best x. The best x is guaranteed to be in this range
        int lo = 1;
        int hi = n;
        while (lo <= hi) {
            int mid = lo + (hi-lo)/2;
            if (check(a, mid, twos)) {
                ans = mid;
                lo = mid+1;
            } else {
                hi = mid-1;
            }
        }
        System.out.println(ans);
    }
 
    // this function checks whether I can use the twos I have to make
    // at least x numbers in the array greater than or equal to x
    private static boolean check(int[] a, int x, int twos) {
        int n = a.length;
        // I only work on the x biggest numbers in the array because they need the
        // least multiplications by 2 to become greater than or equal to x
        int i = n-x;
        while (i < n) {
            int val = a[i];
            while (val < x && twos > 0) {
                val *= 2;
                twos--;
            }
            if (val < x) break;
            i++;
        }
        return i == n;
    }