#include<bits/stdc++.h>
#define inf 1e9
#define eps 1e-6
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define REP(i,a,b) for(int i=a;i>=b;i--)
#define db double
#define ve vector<int>
#define pa pair<int,int>
#define fr first
#define sd second
#define pb push_back
#define mp make_pair
#define MEM(a) memset(a,0,sizeof(a))
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
inline ll read()
{
	char ch = getchar();
	ll s = 0, w = 1;
	while (ch < '0' || ch > '9') {if (ch == '-')w = -1; ch = getchar();}
	while (ch >= '0' && ch <= '9') {s = s * 10 + ch - '0'; ch = getchar();}
	return s * w;
}
#define N 200010
#define ls (x << 1)
#define rs (x << 1 | 1)
#define mid ((l + r) >> 1)
int n, l[N], r[N], m, x, c[N];
vector<int>p[N];
int ans[N], vst[N];
int tag[N << 2];
void add(int x) {for (; x <= n; x += x & (-x))c[x] ++;}
int qry(int x) {int s = 0; for (; x; x -= x & (-x))s += c[x]; return s;}
void clr(int l, int r, int x)
{
	tag[x] = 0;
	if (l == r)return ;
	clr(l, mid, ls); clr(mid + 1, r, rs);
}
void upd(int l, int r, int x, int ql, int qr)
{
	if (l == ql && r == qr) {tag[x] = 1; return ;}
	if (qr <= mid)upd(l, mid, ls, ql, qr);
	else if (ql > mid)upd(mid + 1, r, rs, ql, qr);
	else upd(l, mid, ls, ql, mid), upd(mid + 1, r, rs, mid + 1, qr);
}
int qry(int l, int r, int x, int pos)
{
	if (l == r) {return tag[x];}
	if (tag[x])return 1;
	if (pos <= mid)return qry(l, mid, ls, pos);
	else return qry(mid + 1, r, rs, pos);
}
void sol()
{
	clr(1, n, 1);
	FOR(i, 1, n)c[i] = vst[i] = 0, p[i].clear();
	FOR(i, 1, m)p[r[i]].push_back(l[i]);
	add(x); vst[x] = 1;
	FOR(r, x + 1, n)
	{
		int fl = 0;
		if (p[r].size())vst[r] = 1;
		for (int l : p[r])if (qry(r) - qry(l - 1))
			{
				fl = 1;
				upd(1, n, 1, l, r);
			}
		if (fl)add(r);
	}
	FOR(r, x + 1, n)vst[r] = vst[r] & qry(1, n, 1, r);
}
signed main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	n = read(), m = read(), x = read();
	FOR(i, 1, m)l[i] = read(), r[i] = read();
	sol(); FOR(i, x + 1, n)ans[i] = vst[i];
	x = n - x + 1;
	FOR(i, 1, m)l[i] = n - l[i] + 1, r[i] = n - r[i] + 1;
	FOR(i, 1, m)swap(l[i], r[i]);
	sol();
	// FOR(i, 1, n)cout << vst[i] << ' '; cout << endl;
	FOR(i, 1, (n - x + 1) - 1)ans[i] = vst[n - i + 1];
	FOR(i, 1, n)if (ans[i])cout << i << ' '; cout << '\n';
	return 0;
}