#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef pair<int, int> PII;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define MOD 1000000007
#define INF (1LL<<25)     //33554432
#define PI 3.14159265359
#define EPS 1e-12
//#define int ll

//二分累乗法 xのe乗
ll binpow(ll x, ll e)
{
  ll a = 1, p = x;
  while(e > 0) {
    if(e%2 == 0) {p = (p*p) % MOD; e /= 2;}
    else {a = (a*p) % MOD; e--;}
  }
  return a % MOD;
}

signed main(void)
{
  string n;
  ll ans = 0;
  cin >> n;
  vector<string> v;
  FOR(i, 1, 100000) {
    string s, t;
    char b[1010];
    sprintf(b, "%d", i);
    s = b; t = s;
    reverse(ALL(t));
    v.push_back(s+t);
    v.push_back(s+t.substr(1));
  }
  //cout << "b" << endl;
  sort(ALL(v));
  v.erase(unique(ALL(v)), v.end());
  //cout << "a" << endl;
  for(string i: v) {
    if(i.size() > 9) continue;
    string a(9-i.size(), '0');
    string r = i + a + i;
    if(r.size() < n.size() || (r.size() == n.size() && r <= n)) ans++;
  }
  cout << ans << endl;
  return 0;
}