#include<bits/stdc++.h>
#define all(x) (x).begin(),(x).end()
typedef long long ll;
#define rep(i,n) for(ll i=0, i##_len=(n); i<i##_len; ++i)
#define REP(i,num,n) for(ll i=num, i##_len=(n); i<i##_len; ++i)
#define repprev(i,a,b) for(ll i=b-1;i>=a;i--)
#define reprev(i,n) repprev(i,0,n)
using namespace std;
#define sz(x) ((int)(x).size())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define MEMSET(v, h) memset((v), h, sizeof(v))
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template<class T> int former(const vector<T> &v, T x){
   return upper_bound(v.begin(),v.end(),x) - v.begin() - 1;
}
template<class T> int latter(const vector<T> &v, T x){
    return lower_bound(v.begin(),v.end(),x) - v.begin();
}
#define pb push_back
#define mp make_pair
#define y0 y3487465
#define y1 y8687969
#define j0 j1347829
#define j1 j234892
#define BIT_FLAG_0 (1<<0) // 0000 0000 0000 0001
#define BIT_FLAG_1 (1<<1) // 0000 0000 0000 0010
#define BIT_FLAG_2 (1<<2) // 0000 0000 0000 0100
#define BIT_FLAG_3 (1<<3) // 0000 0000 0000 1000
#define BIT_FLAG_4 (1<<4) // 0000 0000 0001 0000
#define BIT_FLAG_5 (1<<5) // 0000 0000 0010 0000
#define BIT_FLAG_6 (1<<6) // 0000 0000 0100 0000
#define BIT_FLAG_7 (1<<7) // 0000 0000 1000 0000
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
const ll LLINF = 1LL<<60;
const int INTINF = 1<<29;
 
const int MAX = 510000;
const int MOD = 1000000007;
 
long long fac[MAX], finv[MAX], inv[MAX];
 
// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
 
// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
 
struct UnionFind {
    vector<ll> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2
 
    UnionFind(ll n) : par(n, -1) { }
    void init(ll n) { par.assign(n, -1); }
 
    ll root(ll x) {
        if (par[x] < 0) return x;
        else return par[x] = root(par[x]);
    }
 
    bool issame(ll x, ll y) {
        return root(x) == root(y);
    }
 
    bool merge(ll x, ll y) {
        x = root(x); y = root(y);
        if (x == y) return false;
        if (par[x] > par[y]) swap(x, y); // merge technique
        par[x] += par[y];
        par[y] = x;
        return true;
    }
 
    ll size(ll x) {
        return -par[root(x)];
    }
};
 
template <typename T>
vector<T> dijkstra(int s,vector<vector<pair<int, T> > > & G){
   const T INF = numeric_limits<T>::max();
   using P = pair<T, int>;
   int n=G.size();
   vector<T> d(n,INF);
   vector<int> b(n,-1);
   priority_queue<P,vector<P>,greater<P> > q;
   d[s]=0;
   q.emplace(d[s],s);
   while(!q.empty()){
     P p=q.top();q.pop();
     int v=p.second;
     if(d[v]<p.first) continue;
     for(auto& e:G[v]){
       int u=e.first;
       T c=e.second;
       if(d[u]>d[v]+c){
         d[u]=d[v]+c;
         b[u]=v;
         q.emplace(d[u],u);
       }
     }
   }
   return d;
}
 
vector<vector<int> > bfs(vector<string> &s,int sy,int sx,char wall,int dir){
   int h=s.size(),w=s.front().size();
   vector<vector<int> > dp(h,vector<int>(w,-1));
   using P = pair<int, int>;
   queue<P> q;
 
   dp[sy][sx]=0;
   q.emplace(sy,sx);
 
   int dy[]={1,-1,0,0,1,1,-1,-1};
   int dx[]={0,0,1,-1,1,-1,1,-1};
   auto in=[&](int y,int x){return 0<=y&&y<h&&0<=x&&x<w;};
 
   while(!q.empty()){
     int y,x;
     tie(y,x)=q.front();q.pop();
     for(int k=0;k<dir;k++){
       int ny=y+dy[k],nx=x+dx[k];
       if(!in(ny,nx)||s[ny][nx]==wall) continue;
       if(~dp[ny][nx]) continue;
       dp[ny][nx]=dp[y][x]+1;
       q.emplace(ny,nx);
     }
   }
 
   return dp;
}
 
const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
 
int main(void){
    int n, m;
    cin >> n >> m;
    vector<int> c(m);
    rep(i,m) cin >> c[i];
    int dp[n+1];
    rep(i, n + 1){
             dp[i] = INTINF;
    }
    dp[0] = 0;
    rep(i, m){
        for(int j = 0; j + c[i] <= n; ++j){
             dp[j + c[i]] = min(dp[j + c[i]], dp[j] + 1);
       }
    }
 
    cout << dp[n] << endl;
}


Measured memory_bytes: 3280