#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <queue>

using namespace std;
using ull = unsigned long long;
using ll = long long;
using lli = long long int;
using ld = long double;
using pa = pair<int,int>;
#define SORT(v, n) sort(v, v+n);
#define ALL(v) v.begin(),v.end()
#define VSORT(v) sort(ALL(v))
#define GRESORT(v) sort(ALL(v),greater<ll>())
#define REVERSE(v) reverse(ALL(v))
#define overlap(v) v.erase(unique(ALL(v)),v.end())
#define debug(x) cout << #x << ": " << x << endl
#define FOR(i,a,b) for(int i = (a); i < (b); i++)
#define rep(i,n) FOR(i,0,n)
#define RFOR(i,a,b) for(int i = (b-1); i >= a; i--)
#define rrep(i,n) RFOR(i,0,n)
#define INF 999999999
#define mins(a,b) a = min(a,b)
#define maxs(a,b) a = max(a,b)

//4近傍
int dy[]={0, 1, 0, -1};
int dx[]={1, 0, -1, 0};

//8近傍
//int dy[]={0,0,1,-1,1,1,-1,-1};
//int dx[]={1,-1,0,0,1,-1,1,-1};

//ll mod = 1000000007;
ll mod = 1000003;

// 繰り返し2乗法
ll pos(ll x, ll n){
    ll res = 1;
    if(n > 0){
        res = pos(x,n/2);
        if(n%2 == 0) res = (res*res) % mod;
        else res = (((res*res)%mod)*x)%mod;
    }
    return res;
}

void solve() {
    ll x,n; cin >> x >> n;
    vector<ll> a(n);
    rep(i,n) cin >> a[i];
    VSORT(a);
    ll ans = 0;
    rep(i,n){
        ans += pos(x,a[i]);
        ans %= mod;
    }
    cout << ans << endl;
}

int main(){
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);

    solve();

    return 0;
}