#include <bits/stdc++.h>
#define il inline
using namespace std;
typedef long long ll;
const int N=302;
int n,m,mod,f[N][2][N/2][N/2],ans;
il void add(int &x,int y){x=(x+y>=mod?x+y-mod:x+y);}
il int calc(int l,int r){
    r=min(r,n);
    if(l>1 && r<n) return max(0,(r-l+1)/2-1);
    else return max(0,(r-l)/2);
}

int a[N],b[N],g[N][2];
map<vector<int>,int> mp;
vector<int> e;
il void ck(){
    e.clear();
    for(int i=1;i<=n;++i){
        for(int j=1;j<=n;++j){
            g[j][0]=max(g[j-1][0],g[j-1][1]);
            g[j][1]=g[j-1][0]+(i!=j)*a[j];
        }e.push_back(max(g[n][0],g[n][1]));
    }mp[e]=1;
}
void dfs(int x){
    if(x>n){ck();return ;}
    for(int i=0;i<=m;++i) a[x]=i,dfs(x+1);
}
int main(){
    //freopen("c1.in","r",stdin);
    scanf("%d%d%lld",&n,&m,&mod);
    if(n<=6 && m<=6){
        dfs(1);
        printf("%d",mp.size());
        exit(0);
    }

    ans=(n/2ll)+1ll;
    f[1][0][0][0]=f[1][1][1][0]=1ll;
    for(int i=2;i<=n;++i){
        //case 1: a[i]=0
        //(1):a[1]...a[i]=0
        f[i][0][0][i/2]=1ll;
        //(2) exist 1:find the last 1
        for(int j=1;j<=i-1;++j)
            for(int k=1;k*2<=j+1;++k)
                for(int l=0;l+k<=j && l*2<=j;++l)
                    add(f[i][0][k][l+calc(j+1,i+1)],f[j][1][k][l]);
                   // if(i==4 && j==1 && k==1 && l==0) {printf("*** %lld %lld\n",f[j][1][k][l],f[i][0][k][l+calc(]);}

        //case 2: a[i]=1
        //(1) for all j<i j=i(mod 2),a[j]=1
        f[i][1][(i+1)/2][0]=1ll;
        //(2) exist j<i,j=i(mod 2) ,a[j]=0
        //j:the last 1 such that (j+1) ~ i = 1010...101
        for(int j=i-2;j>0;j-=2)
            for(int k=0;k*2<=j;++k)
                for(int l=0;l+k<=j && l*2<=j;++l)
                    add(f[i][1][k+(i-j)/2][l],f[j][0][k][l]);
    }
    for(int mi=0;mi<=(n/2);++mi)
        for(int i=1;i<n && i*2<=n+1;++i){
            if(i-1>mi) break;
            for(int j=0;i+j<=n && j*2<=n;++j){
                if(i+j<mi+1) continue;
                add(ans,f[n][0][i][j]),add(ans,f[n][1][i][j]);
                //printf("%d %d %d %d %lld\n",n,0,i,j,f[n][0][i][j]);
                //printf("%d %d %d %d %lld\n",n,1,i,j,f[n][1][i][j]);
            }
        }

    printf("%d",ans);
    return 0;
}
/*
1,4,6,12,19,35,58,102,172,296,501
2,2,3,3,4,4,5,5,6,6
0,2,3,9,15,31,53,97,290,495
*/