// Original C++ solution (slower):
#include <iostream>

#include <algorithm>

#include <cstdio>

#include <cmath>

#include <cstring>

#define ll long long

#define INF 1000000007

using namespace std;



const int MAXN=1e5+5;



int n,l,r;

ll cnt,sum;

int a[MAXN],b[MAXN],c[MAXN];



void work(int begin,int end){

	if(begin==end)

		return;

	int mid=(begin+end)/2;

	work(begin,mid);

	work(mid+1,end);

	int i=begin,j=mid+1,k=begin;

	while(i<=mid && j<=end){

		if(b[i]<b[j])

			c[k++]=b[i++];

		else{

			cnt+=(i-begin);

			c[k++]=b[j++];

		}

	}

	while(i<=mid)

		c[k++]=b[i++];

	while(j<=end){

		cnt+=(i-begin);

		c[k++]=b[j++];

	}

	for(int l=begin;l<=end;l++)

		b[l]=c[l];

}



bool check(int m){

	cnt=0;

	for(int i=1;i<=n;i++){

		if(a[i]<=m)

			b[i]=1;

		else

			b[i]=-1;

	}

	for(int i=1;i<=n;i++){

		b[i]+=b[i-1];

	}

	work(0,n);

	if(cnt>=(sum/2+1))

		return true;

	return false;

}



int main(){

	scanf("%d",&n);

	l=INF;r=0;sum=0;

	for(int i=1;i<=n;i++){

		scanf("%d",&a[i]);

		l=min(l,a[i]);

		r=max(r,a[i]);

		sum+=i;

	}

	while(l<r){

		int m=(l+r)/2;

		if(check(m)){

			r=m;

		}else{

			l=m+1;

		}

	}

	printf("%d\n",r);

	return 0;

}




// Optimized solution to the same problem (faster):
#include<stdio.h>

#include<algorithm>

using namespace std;

#define fo(i,a,b) for(int i=a;i<=b;i++)

#define fd(i,a,b) for(int i=a;i>=b;i--)

int n,a[110000],l,r,mi,f[220000];

long long m;

inline void add(int x){

	while (x<=n+n){

		f[x]++;

		x+=x&(-x);

	}

}

inline long long ask(int x){

	long long ans=0;

	while (x){

		ans+=f[x];

		x-=x&(-x);

	}

	return ans;

}

inline bool che(int x){

	int s=n+1;long long ans=0;//s=n TLE

	fo(i,0,n+n) f[i]=0;

	add(s);

	fo(i,1,n){

		if (a[i]>=x) s++;else s--;

		ans+=ask(s);

		add(s);

	}

	return ans>=m;

}

int main(){

	scanf("%d",&n);

	fo(i,1,n) scanf("%d",&a[i]);

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

	m=(m+1)/2;

	l=0x7fffffff;r=0;

	fo(i,1,n){

		if (a[i]<l) l=a[i];

		if (a[i]>r) r=a[i];

	}

	l--;r++;

	while (l+3<r){

		mi=(l+r)>>1;

		if (che(mi)) l=mi+1;else r=mi-1;

	}

	fd(i,r+1,l-1) if (che(i)){//che(mi)!!!

		printf("%d\n",i);

		return 0;

	}

	return 0;

}

// Fastest known solution for this problem:
#define _USE_MATH_DEFINES



#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <iostream>

#include <cmath>

#include <algorithm>

#include <vector>

#include <queue>

#include <map>

#include <list>

#include <stack>



using namespace std;



typedef pair<long long int, long long int> P;



long long int INF = 1e18;

long long int MOD = 1e9 + 7;



long long int N, sz = 150000;

long long int A[110000];

bool flag[110000];

long long int BIT[210000];



void add(int a, long long int w){

	

	// a番目の要素にwを足す O(log N)

	

	for(int x = a; x <= sz; x += x & -x){

		BIT[x] += w;

	}

}



long long int sum(int a){

	

	// 1からaまでの要素の合計を返す O(log N)

	

	long long int ret = 0;

	for(int x = a; x > 0; x -= x & -x){

		ret += BIT[x];

	}

	return ret;

}



void init(){

	for(int i = 0; i < sz; i++){

		BIT[i] = 0;

	}

}



bool check(long long int min_A){

	for(int i = 0; i < N; i++){

		flag[i] = (A[i] >= min_A);

	}

	vector<P> v;

	long long int cnt = 0, S = 0;

	v.push_back(make_pair(0, 1));

	for(int i = 0; i < N; i++){

		if(flag[i]){

			S += 1;

		}else{

			S -= 1;

		}

		v.push_back(make_pair(S, i + 2));

	}

	sort(v.begin(), v.end());

	init();

	for(int i = 0; i < v.size(); i++){

		cnt += sum(v[i].second);

		add(v[i].second, 1);

	}

	//cout << min_A << " " << cnt << endl;

	return cnt >= (N * (N + 1) / 2 + 1) / 2;

}



long long int bin_search(long long int lef, long long int rig){

	long long int middle = (lef + rig) / 2;

	//cout << middle << " ";

	if(rig - lef <= 1){

		if(check(rig)){

			return rig;

		}else{

			return lef;

		}

	}

	if(check(middle)){

		lef = middle;

	}else{

		rig = middle;

	}

	return bin_search(lef, rig);

}



int main(){

	

	cin >> N;

	

	for(int i = 0; i < N; i++){

		cin >> A[i];

	}

	

	cout << bin_search(0, 1e10) << endl;

	

	return 0;

}