#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
ll n,m,sum;
ll a[15000];
ll huiwen(ll n)//倒过来和的数与原来相等，那么就是回文数
{
    ll ans=n,t=n;//ll ans=n 若不分偶数位，则为ans=0;
    while(t)
    {
        ans=ans*10+t%10;
        t/=10;
    }
    return ans;
}

int main()
{
    cin>>n>>m;
    for(ll i=1;i<=n;i++)
    {
        sum=(sum+huiwen(i))%m;
    }
    printf("%lld\n",sum%m);
}

#include<bits/stdc++.h>
#define IO ios::sync_with_stdio(false);\
    cin.tie(0);\
    cout.tie(0);
using namespace std;
const int maxn = 1e5+10;
typedef long long LL;
typedef pair<int,int> P;
LL k,p;

LL getnow(LL n)
{
    char str[20] = {0};
    sprintf(str,"%lld",n);
    int len = strlen(str);
    for(int i=len;i<len+len;i++)
        str[i] = str[2*len-i-1];
    LL cnt = 0;
    sscanf(str,"%lld",&cnt);
    return cnt;
}

void solve()
{
    LL ans = 0;
    for(int i=1;i<=k;i++)
        ans = (ans+getnow(i)) % p;
//    cout<<getnow(k)<<endl;
    cout<<ans<<endl;
}

int main()
{
//    IO;
    cin>>k>>p;
    solve();
    return 0;
}