/*ver 6*/
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <iomanip>
#include <vector>
#include <string>
#include <utility>
#include <algorithm>
#include <queue>
#include <tuple>
#include <map>
#include <list>
#include <set>
using namespace std;
void init() {cin.tie(0);ios::sync_with_stdio(false);cout << fixed << setprecision(15);}
using ll = long long;
using ld = long double;
using vl = vector<ll>;
using vd = vector<ld>;
using vs = vector<string>;
using vb = vector<bool>;
using vvl = vector<vector<ll>>;
using vvd = vector<vector<ld>>;
using vvs = vector<vector<string>>;
using vvb = vector<vector<bool>>;
using pll = pair<ll,ll>;
using mll = map<ll,ll>;
#define EACH(x,v) for(auto& x : v)
#define FOR(i,a,b) for(ll i=(ll)a;i<(ll)b;i++)
#define REFOR(i,a,b) for(ll i=(ll)b-1;i>=(ll)a;i--)
#define REP(i,n) for(ll i=0;i<(ll)n;i++)
#define REREP(i,n) for(ll i=(ll)n-1;i>=0;i--)
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define mp make_pair
const ll INF = 1LL << 60;
#define CLR(mat,f) memset(mat, f, sizeof(mat))
#define IN(a, b, x) (a<=x&&x<=b)
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() ) //被り削除
#define debug cout << "line : " << __LINE__ << " debug" << endl;
#define ini(...) int __VA_ARGS__; in(__VA_ARGS__)
#define inl(...) long long __VA_ARGS__; in(__VA_ARGS__)
#define ind(...) long double __VA_ARGS__; in(__VA_ARGS__)
#define ins(...) string __VA_ARGS__; in(__VA_ARGS__)
#define inc(...) char __VA_ARGS__; in(__VA_ARGS__)
void in(){}
template <typename T,class... U> void in(T &t,U &...u){ cin >> t; in(u...);}
template <typename T> void in1(T &s) {REP(i,s.size()){in(s[i]);}}
template <typename T> void in2(T &s,T &t) {REP(i,s.size()){in(s[i],t[i]);}}
template <typename T> void in3(T &s,T &t,T &u) {REP(i,s.size()){in(s[i],t[i],u[i]);}}
template <typename T> void in4(T &s,T &t,T &u,T &v) {REP(i,s.size()){in(s[i],t[i],u[i],v[i]);}}
template <typename T> void in5(T &s,T &t,T &u,T &v,T &w) {REP(i,s.size()){in(s[i],t[i],u[i],v[i],w[i]);}}
void out(){cout << endl;}
template <typename T,class... U> void out(const T &t,const U &...u){cout << t; if(sizeof...(u)) cout << " "; out(u...);}
void die(){cout << endl;exit(0);}
template <typename T,class... U> void die(const T &t,const U &...u){cout << t; if(sizeof...(u)) cout << " "; die(u...);}
template <typename T> void outv(T s) {REP(i,s.size()){if(i!=0)cout<<" ";cout<<s[i];}cout<<endl;}
template <typename T> void out1(T s) {REP(i,s.size()){out(s[i]);}}
template <typename T> void out2(T s,T t) {REP(i,s.size()){out(s[i],t[i]);}}
template <typename T> void out3(T s,T t,T u) {REP(i,s.size()){out(s[i],t[i],u[i]);}}
template <typename T> void out4(T s,T t,T u,T v) {REP(i,s.size()){out(s[i],t[i],u[i],v[i]);}}
template <typename T> void out5(T s,T t,T u,T v,T w) {REP(i,s.size()){out(s[i],t[i],u[i],v[i],w[i]);}}
#define all(v) (v).begin(),(v).end()
#define rall(v) (v).rbegin(),(v).rend()
template <typename T> T allSum(vector<T> a){T ans=T();EACH(it,a)ans+=it;return ans;}
ll ceilDiv(ll a,ll b) {return (a+b-1)/b;}
ld dist(pair<ld,ld> a, pair<ld,ld> b){return sqrt(abs(a.fi-b.fi)*abs(a.fi-b.fi)+abs(a.se-b.se)*abs(a.se-b.se));} // ２点間の距離 
ll gcd(ll a, ll b) { return b != 0 ? gcd(b, a % b) : a; }
ll lcm(ll a,ll b){ return a / gcd(a,b) * b;}
template <class A, class B> inline bool chmax(A &a, const B &b) { return b > a && (a = b, true); }
template <class A, class B> inline bool chmin(A &a, const B &b) { return b < a && (a = b, true); }
#define YES(n) ((n) ? "YES" : "NO"  )
#define Yes(n) ((n) ? "Yes" : "No"  )
#define yes(n) ((n) ? "yes" : "no"  )

int main(){
	init();
	inl(a,b);
	out((a+b)/2,(a-b)/2);
	return 0;
}
#include <bits/stdc++.h>
 
using namespace std;
 
/*
Fix k.
For this k, we will:
Compute the maximum possible sum of k items that is equal to X modulo k.
We will do this by using a DP approach. We go through the items one by one and decide IN ORDER whether we use them.
Our DP state is (# items used, current sum % k, items gone through) -> value.
For each k, we have 10^6 DP states.
The processing for each state is constant time.
So we have at most 10^8 "things" to do.
*/
 
void maxeq(long long& a, long long b){
	a = max(a, b);
}
 
long long DP[102][101][101];
int main(){
	long long n, x;
	cin>>n>>x;
	vector<long long> A(n);
	for (int i = 0; i < n; i++) cin>>A[i];
	long long ans = 2e18;
	// (# items used, current sum % k, items gone through) -> value.
	for (int k = 1; k <= n; k++){
		//long long DP[k+1][k][n+1];
		for (int a = 0; a <= k+1; a++){
			for (int b = 0; b < k; b++){
				for (int c = 0; c <= n; c++){
					DP[a][b][c] = LLONG_MIN / 3;
				}
			}
		}
		DP[0][0][0] = 0;
		for (int c = 0; c < n; c++){ // we decide whether to take item c
			for (int a = 0; a < k+1; a++){
				for (int b = 0; b < k; b++){
					// take
					maxeq(DP[a + 1][(b + A[c]) % k][c + 1], DP[a][b][c] + A[c]);
					// don't take	
					maxeq(DP[a][b][c + 1], DP[a][b][c]);
				}	
			}
		}
		if (DP[k][x % k][n] > 0) ans = min(ans, (x - DP[k][x % k][n]) / k);
	}
	cout << ans << endl;
}