// Original C++ solution (slower):
#include<bits/stdc++.h>

//CLOCKS_PER_SEC

#define se second

#define fi first

#define ll long long

#define Pii pair<int,int>

#define Pli pair<ll,int>

#define ull unsigned long long

#define pb push_back

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)

const int N=1e5+10;

const int INF=0x3f3f3f3f;

const int mod=1e9+7;

using namespace std;

int n,val[N<<1];

int a[N],b[N];

int sum[N];

void add(int x){

	while(x<N*2){

		val[x]+=1;

		x+=x&-x;

	}

}

ll get(int x){

	ll ans=0;

	while(x>0){

		ans+=val[x];x-=x&-x;

	}

	return ans;

}

bool check(int x){

	for(int i=1;i<=n;++i)sum[i]=sum[i-1]+(a[i]>=x?1:-1);

	ll tot=0;

	memset(val,0,sizeof(val));

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

		tot+=get(sum[i]+N);

		add(sum[i]+N);

	}

	return tot>=1LL*n*(n+1)/4;

}

int main(){

	scanf("%d",&n);

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

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

	}

	sort(b+1,b+1+n);

	int l=1,r=n;

	int ans=1;

	while(l<=r){

		int mid=(l+r)>>1;

		if(check(b[mid])){

			ans=mid;

			l=mid+1;

		}

		else r=mid-1;

	}

	cout<<b[ans]<<endl;

	return 0;

}  

// Optimized solution to the same problem (faster):
#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<cmath>

#include<cstring>

#include<algorithm>

#include<queue>

#define inl inline

#define re register

#define LL long long

#define MAXN 10100

using namespace std;

struct con

{

	int p;LL sum;

}hh[100100];

LL read()

{

    LL f=1,p=0;

    char c=getchar();

    while(c>'9'||c<'0'){if(c=='-')f=-1;c=getchar();}

    while(c>='0'&&c<='9'){p=p*10+c-'0';c=getchar();}

    return f*p;

}

void write(LL x)

{

    if(x<0)putchar('-'),x=-x;

    if(x/10)write(x/10);

    putchar(x%10+'0');

}

LL n,a[100100];

LL l=1000000000l,r,mid,ans,f[100100];

LL sum[100100];

inl LL lowbit(LL x){return x&(-x);}

inl bool cmp(const con&a,const con &b)

{

	if(a.sum==b.sum)return a.p<b.p;

	return a.sum<b.sum;

}

void add(int b){for(int i=b;i<=n;i+=lowbit(i))f[i]++;}

LL find(int a)

{

	LL u=0;

	for(int i=a;i>0;i-=lowbit(i))u+=f[i];

	return u;

}

bool check()

{

	ans=0;

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

	{

		if(a[i]<mid)hh[i].sum=-1;

		else hh[i].sum=1;

	}

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

		hh[i].sum+=hh[i-1].sum;

	for(int i=1;i<=n;i++)hh[i].p=i;

	sort(hh+1,hh+n+1,cmp);

	memset(f,0,sizeof(f));

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

	{

		add(hh[i].p);

		ans+=find(hh[i].p-1);

		if(hh[i].sum>=0)ans++; 

	}

	return 4*ans <n*(n+1);

}

int main()

{

	n=read();

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

	{

		a[i]=read();

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

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

	}

	while(l<r)

	{

		mid=(l+r+1)/2;

		if(check()) r=mid-1;

		else l=mid;

	}

	write(l);

	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;

}