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

#include<cstring>

#include<cstdio>

#define ll long long

using namespace std;

const int N=100000+7;

int n,a[N],l,r,mid,c[3*N],t[N];

ll s[N];

int lowbit(int x){

	return x&-x;

}

void add(int x){

	for(int i=x;i<=2*N;i+=lowbit(i))

		c[i]++;

}

int query(int x){

	ll sum=0;

	for(int i=x;i>=1;i-=lowbit(i))

		sum+=c[i];

	return sum;

}

bool check(int now){

    for(int i=1;i<=2*N;i++)c[i]=0;		//用memset会出错 

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

		s[i]=s[i-1]+(a[i]>=now ?1 :-1);

		//if(now==9)printf("%d ",s[i]);

	}

	//printf("\n");

	ll sum=0;

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

		sum+=query(s[i]+N);

		add(s[i]+N);

	}

	//printf("%d %d\n",now,sum);

	return sum>=1ll*n*(n+1)/4;

}

int main(){

	//freopen("1.txt","r",stdin);

	cin>>n;

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

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

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

	}

	while(l<=r){			//一定要用 <=，不然会漏情况 

		mid=(l+r)>>1;

		if(check(mid))l=mid+1;

		else r=mid-1;

	}

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

	return 0;

}

/*

3

10 30 20



*/

/*

10

5 9 5 9 8 9 3 5 4 3



*/

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

using namespace std;



const int N = 110000;



int a[N], sum[N];

int BIT[3 * N];



int main() {

	

	ios::sync_with_stdio(0);

	cin.tie(0), cout.tie(0);

	

	int n; cin >> n;

	for (int i = 1; i <= n; i++) cin >> a[i];

	

	int st = 0, en = 1e9, mid, ans = en;

	

	auto add = [&](int u, int v) {

		for (int i = u; i <= 3 * (n + 1); i += i & -i) BIT[i] += v;

	};

	

	auto get = [&](int u) {

		int ret = 0;

		for (int i = u; i > 0; i -= i & -i) ret += BIT[i];

		return ret;

	};

	

	auto can = [&](int mid) {

		sum[0] = 0;

		for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + 2 * (a[i] <= mid);

		for (int i = 0; i <= 3 * (n + 1); i++) BIT[i] = 0;

		long long cnt = 0;

		add(0 + n + 1, 1);

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

			cnt += get(sum[i] - i + n);

			add(sum[i] - i + n + 1, 1);

		}

		return cnt * 2 > (long long)n * (n + 1) / 2;

	};

	

	while (st <= en) {

		mid = st + en >> 1;

		if (can(mid)) ans = mid, en = mid - 1;

		else st = mid + 1;

	}

	cout << ans << "\n";

	

	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;

}