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
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<int, pii> pipii;
typedef pair<pii, pii> piipii;

#define all(a) (a).begin(), (a).end()
#define sz(a) (int)(a).size()
#define eb emplace_back
int a[100005], b[100005], sum[100005];

int x = 100002;
int ft[200005];
void update(int i, int v){
    for(;i<=200004;i+=(i&-i)) ft[i] += v;
}
int query(int i){
    int res = 0;
    for(;i>0;i-=(i&-i)) res+=ft[i];
    return res;
}

int main(){
    int n;
    scanf("%d", &n);
    for(int i=1;i<=n;i++) scanf("%d", &a[i]);

    ll m = n*1ll*(n+1)/2;
    m = m/2 + 1;

    int lo = 1, hi = 1e9, ans = 0;
    while(lo <= hi){
        int mid = (lo+hi)/2;

        memset(ft, 0, sizeof(ft));
        for(int i=1;i<=n;i++){
            if(a[i] <= mid) b[i] = 1;
            else b[i] = -1;
            sum[i] = sum[i-1] + b[i];
        }

        ll cnt = 0;
        update(x, 1);
        for(int i=1;i<=n;i++){
            cnt += query(x+sum[i]-1);
            update(x+sum[i], 1);
        }

        if(cnt >= m) ans = mid, hi = mid-1;
        else lo = mid+1;
    }

    printf("%d\n", ans);
}