#include<bits/stdc++.h>
using namespace std;
//#pragma GCC optimize("Ofast")

#define rep(i,n) for(ll i=0;i<n;i++)
#define repl(i,l,r) for(ll i=(l);i<(r);i++)
#define per(i,n) for(ll i=(n)-1;i>=0;i--)
#define perl(i,r,l) for(ll i=r-1;i>=l;i--)
#define fi first
#define se second
#define pb push_back
#define ins insert
#define pqueue(x) priority_queue<x,vector<x>,greater<x>>
#define all(x) (x).begin(),(x).end()
#define CST(x) cout<<fixed<<setprecision(x)
#define vtpl(x,y,z) vector<tuple<x,y,z>>
#define rev(x) reverse(x);
using ll=long long;
using vl=vector<ll>;
using vvl=vector<vector<ll>>;
using pl=pair<ll,ll>;
using vpl=vector<pl>;
using vvpl=vector<vpl>;
const ll MOD=1000000007;
const ll MOD9=998244353;
const int inf=1e9+10;
const ll INF=4e18;
//const ll dy[9]={1,0,-1,0,1,1,-1,-1,0};
//const ll dx[9]={0,1,0,-1,1,-1,1,-1,0};
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}


//実行時modint 
ll mod;
int max_n = 200005;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-()  { return mint(-x);}
  mint& operator+=( mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=( mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=( mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+( mint a) { return mint(*this) += a;}
  mint operator-( mint a) { return mint(*this) -= a;}
  mint operator*( mint a) { return mint(*this) *= a;}
  mint pow(ll t) {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }
  bool operator==( mint &p) { return x == p.x; }
  bool operator!=( mint &p) { return x != p.x; }
  // for prime mod
  mint inv() { return pow(mod-2);}
  mint& operator/=(mint a) { return *this *= a.inv();}
  mint operator/( mint a) { return mint(*this) /= a;}
};
istream& operator>>(istream& is, mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, mint& a) { return os << a.x;}
using vm=vector<mint>;
using vvm=vector<vm>;

//https://atcoder.jp/contests/abc230/submissions/44985883
vector<tuple<ll,ll,ll>> quotient_floors(ll n){
    //floor(n/i)=xを満たすx,i∈[l,r)をこの順に返す。lの昇順、つまりxの降順なことに注意。
    vector<tuple<ll,ll,ll>> ret;
    for(ll l=1;l<=n;){
        ll x=n/l;
        ll r=n/x+1;
        ret.emplace_back(x,l,r);
        l=r;
    }
    return ret;
}


struct osak{
    vector<long long> lpf;// least prime factor
    vector<long long> prime;// prime table
    osak(long long n){//linear_sieve
        lpf=vector<long long>(n+1,-1);
        for (int d = 2; d <= n; ++d) {
          if(lpf[d]==-1){
            lpf[d]=d;prime.emplace_back(d);
          }
          for(auto p:prime){
            if(p*d>n||p>lpf[d])break;
            lpf[p*d]=p;
          }
        }
    }
    map<long long,long long> factor(int n) {
        map<long long,long long> factor;
        while (n > 1) {
            factor[lpf[n]]++;
            n /= lpf[n];
        }
        return factor;
    }

    vector<long long> divisor(int N){//O(div.size())
        map<long long,long long> facs=factor(N);
        vector<long long> ret={1};
        for(auto p:facs){
            ll range=ret.size();
            ll now=1;
            for(int i=0;i<p.se;i++){
                now*=p.fi;
                for(int j=0;j<range;j++){
                    ret.emplace_back(ret[j]*now);
                }
            }
        }
        sort(ret.begin(),ret.end());
        return ret;
    }
};

int main(){
    ll n;cin >> n >> mod;
    //for(auto [p,q,r]:quotient_floors(99))cout << p <<" " << q <<" " << r << endl;
    vm ans(n+1);
    osak os(300010);
    vvm dp(n+1);
    for(ll i=1;i<=n;i++){
        rep(_,n/i+5)dp[i].emplace_back(0);
    }
    for(ll i=3;i<=n;i++){
        mint sum=1;
        mint cof=1;
        mint inv=mint(i).inv();
        for(auto x:os.divisor(i)){
            dp[x][i/x+1]=dp[x][i/x];
            cof-=inv;
        }
        for(auto [x,y,z]:quotient_floors(i)){
            sum+=(dp[x][z]-dp[x][y])/i;
        }
        ans[i]=sum/cof;
        //cout << sum <<" " << cof << endl;
        for(auto x:os.divisor(i)){
            dp[x][i/x+1]+=ans[i];
        }
    }
    cout << ans.back() << endl;

}