#include"altar.h"
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <numeric>
#include <vector>
#include <queue>
#include <map>
#include <cmath>
#include <cctype>
#include <set>
#include <random>

namespace uvu
{
#define LOCAL ____________DONT_DEFINE_ME____________
// #define ll long long
// #define inf 0x3f3f3f3f
#define int long long
#define inf 0x3f3f3f3f3f3f3f3fll
#define infll 0x3f3f3f3f3f3f3f3fll
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define gline debug("now is #%d\n", __LINE__)
#define pii std::pair <int, int>
#define mkp std::make_pair
#define fi first
#define se second
char _ST_;
const int BUFSIZE = (1 << 20);
char ibuf[BUFSIZE], *iS = ibuf, *iT = ibuf;
char obuf[BUFSIZE], *oS = obuf, *oT = obuf + BUFSIZE;
char getc()
{
#ifdef LOCAL
	return getchar();
#else
	if(iS == iT) iT = (iS = ibuf) + fread(ibuf, 1, BUFSIZE, stdin);
	return iS == iT ? EOF : *iS++;
#endif
#define getchar ERR
}

void Flush() { fwrite(obuf, 1, oS - obuf, stdout); oS = obuf; }
struct Flusher { ~Flusher(){ Flush(); } }iamflusher;

void putc(char c)
{
#ifdef LOCAL
	putchar(c);
#else
	*oS++ = c;
	if(oS == oT) Flush();
#endif
#define putchar ERR
}

template <typename T = int> T read()
{
	T x = 0, f = 1; char c = getc();
	for(; !isdigit(c); c = getc()) if(c == '-') f = -1;
	for(;  isdigit(c); c = getc()) x = (x << 3) + (x << 1) + (c ^ 48);
	return x * f;
}

template <typename T> void print(T x, char c)
{
static int sta[BUFSIZE], top;
	top = 0;
	if(x < 0) putc('-'), x = -x;
	if(!x) sta[top = 1] = 0;
	for(; x; x /= 10) sta[++top] = x % 10;
	for(; top; ) putc(sta[top--] ^ 48);
	if(c) putc(c);
}

int readstr(char *s, int base)
{
	int idx = base - 1; char c = getc();
	for(; !(isdigit(c) || isalpha(c) || c == '(' || c == '?'); c = getc());
	for(;   isdigit(c) || isalpha(c) || c == '(' || c == '?' ; c = getc()) s[++idx] = c;
	return idx - base + 1;
}

void printf(const char *s) { for(; *s; s++) putc(*s); }
template <typename T, typename ... Args>
void printf(const char *s, T x, Args ... rest)
{
	for(; *s; s++)
	{
		if(*s != '%') { putc(*s); continue; }
		s++; if(*s == 'd') print(x, 0);
		else if(*s == 'c') putc(x);
		printf(s + 1, rest ...);
		return;
	}
}

template <typename T> void ckmax(T &x, T y) { x = x > y ? x : y; }
template <typename T> void ckmin(T &x, T y) { x = x < y ? x : y; }
#define mod 998244353
// #define mod 1000000007
int sm(int x) { return x >= mod ? x - mod : x; }
void plus_(int &x, int y) { x = (x + y >= mod ? x + y - mod : x + y); }
void mul_(int &x, int y) { x = 1ll * x * y % mod; }
int ksm(int a, int b) { int res = 1; for(; b; b >>= 1, mul_(a, a)) if(b & 1) mul_(res, a); return res; }

// 核心：答案的权值 <= 2，即在一个竞赛图上一定存在一个点到其他所有点距离 <= 2

// 由出度证明：考虑出度最大的点，若它的权值 >= 3，那么第三层的点一定出度严格比它大（1+第二层），与最大矛盾。故出度最大点的权值一定 <= 2
// 归纳法证明：若存在入度 = 0 的点那么它权值 = 1，一定合法。否则对于任意一个 x，已知它的入点构成的导出子图归纳合法，导出子图中合法的那个点到 x 和 x 的出点距离都 <= 2，故它仍然合法。

// 由归纳法，每次随机一个点递归到入点即可，每次期望 / 2。

std::mt19937 rnd(114514);
int n;

int solve(int _n)
{
	// memset(h, idx = -1, sizeof(h));
	n = _n;
	std::vector <int> now, tmp;
	for(int i = 1; i <= n; i++) now.push_back(i);
	while((int)now.size() > 1)
	{
		std::vector <int>().swap(tmp);
		std::shuffle(now.begin(), now.end(), rnd);
		int x = now[0], sz = now.size();
		for(int i = 1; i < sz; i++) if(sense(now[i], x)) tmp.push_back(now[i]);
		if(tmp.empty()) return x;
		std::swap(tmp, now);
	}
	return now[0];
}

#ifdef int
	#undef int
#endif
}

int altar(int n){
	return uvu::solve(n);
}