/*input
11 300
6 8 8 12 0 11 6 1 0 15 5
*/
#include<bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
//#pragma GCC target("avx2")
//order_of_key #of elements less than x
// find_by_order kth element
using ll=long long;
using ld=long double;
using pii=pair<int,int>;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()),c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
const ll maxn=5e3+5;
const ll maxlg=__lg(maxn)+2;
const ll INF64=4e18;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-4;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
template<typename T1,typename T2>
ostream& operator<<(ostream& out,pair<T1,T2> P){
    out<<P.f<<' '<<P.s;
    return out;
}
template<typename T>
ostream& operator<<(ostream& out,vector<T> V){
    REP(i,sz(V)) out<<V[i]<<((i!=sz(V)-1)?"\n":"");
    return out;
}
ll mult(ll a,ll b){
    return a*b%MOD;
}
ll mypow(ll a,ll b){
    a%=MOD;
    if(a==0) return 0;
    if(b<=0) return 1;
    ll res=1LL;  
    while(b){
        if(b&1) res=(res*a)%MOD;
        a=(a*a)%MOD;
        b>>=1;
    }
    return res;
}
ll dp[(1<<13)][13][505];
int arr[15];
int main(){
    ios::sync_with_stdio(false),cin.tie(0);
    int n,m;
    cin>>n>>m;
    REP(i,n){
        cin>>arr[i];
    }
    REP(j,n){
        int ans=0;
        REP(k,n){
            int need=max(0,arr[k]-arr[j]+(j>k));
            MXTO(ans,need);
        }
        dp[(1<<j)][j][ans*n]=1;
    }
    REP1(i,(1<<n)-1){
        int cn=__builtin_popcount(i);
        if(cn<=1) continue;
        REP(j,n){
            if(i&(1<<j)){
                REP(k,n){
                    if(k==j) continue;
                    if((i&(1<<k))){
                        int need=max(0,arr[k]-arr[j]+(j>k));
                        REP(a,m+1){
                            int nw=a+need*(n-cn+1);
                            if(nw>m) break;
                            dp[i][j][nw]+=dp[i^(1<<j)][k][a];
                        }
                    }
                }
                /*REP(a,m+1){
                    if(dp[i][j][a]) cout<<i<<' '<<j<<' '<<a<<' '<<dp[i][j][a]<<'\n';
                }*/
            }
        }
    }
    ll ans=0;
    REP(i,n) REP(j,m+1) ans+=dp[(1<<n)-1][i][j];
    cout<<ans;
}