We will define the median of a sequence b of length M, as follows:

* Let b' be the sequence obtained by sorting b in non-decreasing order. Then, the value of the (M / 2 + 1)-th element of b' is the median of b. Here, / is integer division, rounding down.



For example, the median of (10, 30, 20) is 20; the median of (10, 30, 20, 40) is 30; the median of (10, 10, 10, 20, 30) is 10.

Snuke comes up with the following problem.

You are given a sequence a of length N. For each pair (l, r) (1 \leq l \leq r \leq N), let m_{l, r} be the median of the contiguous subsequence (a_l, a_{l + 1}, ..., a_r) of a. We will list m_{l, r} for all pairs (l, r) to create a new sequence m. Find the median of m.

Constraints

* 1 \leq N \leq 10^5
* a_i is an integer.
* 1 \leq a_i \leq 10^9

Input

Input is given from Standard Input in the following format:


N
a_1 a_2 ... a_N


Output

Print the median of m.

Examples

Input

3
10 30 20


Output

30


Input

1
10


Output

10


Input

10
5 9 5 9 8 9 3 5 4 3


Output

8
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.TreeSet;
 
public class Main {
	static int n;
	static long half;
	static int[] a;
 
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		n = Integer.parseInt(br.readLine());
		String[] sa = br.readLine().split(" ");
		a = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = Integer.parseInt(sa[i]);
		}
		br.close();
 
		half = (long) n * (n + 1) / 4;
		TreeSet<Integer> set = new TreeSet<>();
		for (int i = 0; i < n; i++) {
			set.add(a[i]);
		}
		Integer[] arr = set.toArray(new Integer[0]);
 
		int ok = -1;
		int ng = arr.length;
		while (Math.abs(ok - ng) > 1) {
			int mid = (ok + ng) / 2;
			if (judge(arr[mid])) {
				ok = mid;
			} else {
				ng = mid;
			}
		}
		System.out.println(arr[ok]);
	}
 
	static boolean judge(int med) {
		int[] b = new int[n + 1];
		for (int i = 0; i < n; i++) {
			if (a[i] >= med) {
				b[i + 1] = 1;
			} else {
				b[i + 1] = -1;
			}
		}
		for (int i = 1; i < b.length; i++) {
			b[i] += b[i - 1];
		}
 
		long t = tentousuu(b);
		return t <= half;
	}
 
	static long tentousuu(int[] a) {
		int n = a.length;
		int min = Integer.MAX_VALUE;
		int max = Integer.MIN_VALUE;
		for (int i = 0; i < n; i++) {
			min = Math.min(min, a[i]);
			max = Math.max(max, a[i]);
		}
		min--;
		int[] b = new int[n];
		for (int i = 0; i < n; i++) {
			b[i] = a[i] - min;
		}
 
		BIT bit = new BIT(max - min);
		long ret = 0;
		for (int i = 0; i < n; i++) {
			ret += i - bit.sum(b[i]);
			bit.add(b[i], 1);
		}
		return ret;
	}
 
	static class BIT {
		int n;
		long[] arr;
 
		public BIT(int n) {
			this.n = n;
			arr = new long[n + 1];
		}
 
		void add(int idx, long val) {
			for (int i = idx; i <= n; i += i & -i) {
				arr[i] += val;
			}
		}
 
		long sum(int idx) {
			long sum = 0;
			for (int i = idx; i > 0; i -= i & -i) {
				sum += arr[i];
			}
			return sum;
		}
	}
}