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>
//#define IORD
#define eps 1e-8
#define inf 0x3f3f3f3f
//#define mod (1000000007)

using namespace std;
typedef long long LL;
typedef pair<int,int> pp;

const int N=1e5+10;
const int maxn=N*2+10;
int n,a[N],b[N],ais[N];
LL cnt=0;

int bit[maxn+1];
int lb(int x)
{
    return x&-x;
}

int sum(int i)
{
    int s=0;
    while(i>0)
    {
        s+=bit[i];
        i-=lb(i);
    }
    return s;
}

void add(int i,int x)
{
    while(i<=maxn)
    {
        bit[i]+=x;
        i+=lb(i);
    }
}

int Che(int x)
{
    memset(bit,0,sizeof(bit));
    memset(ais,0,sizeof(ais));
    for(int i=1;i<=n;i++) ais[i]=(ais[i-1]+(a[i]>=x?1:-1));
    cnt=0;
    for(int i=0;i<=n;i++)
    {
        cnt+=sum(ais[i]+N);
        add(ais[i]+N,1);
    }
    return cnt>=1ll*(n+1)*n/4;
}

int Sol()
{
    memcpy(b,a,sizeof(a));
    sort(b+1,b+n+1);
    int l=0,r=n+1,m=(l+r)/2;
    while(l+1<r)
    {
        if(Che(b[m])) l=m;else r=m;
        m=(l+r)/2;
    }
    return b[l];
}

int main()
{
    #ifdef IORD
    freopen("in.txt","r",stdin);
	freopen("out.txt","w",stdout);
	#endif // IORD
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",a+i);
    printf("%d\n",Sol());
    return 0;
}
