#include <bits/stdc++.h>
/*
Ascending = no cost
Descending = cost of double AD (difference from normal differences)
Add regular difference (end - start)?
For every start and end skyscraper, DP over cost (cost <= 1000)?
Inversions are what matters
Iterate over buildings in sorted order
The next building will cause an inversion unless placed at end
Either cause an inversion or change the difference by placing at end
The difference is stored with the inversions
As the new element is inserted, an inversion might be replaced
The old last element is not stored
O(N^2L) ~ 10^7
Count number of permutations where a certain number is "exposed" once/twice
*/
int n,l;
std::vector<int> builds;
int main(){
std::cin>>n>>l;
for(int i=0;i<n;i++){
int x;
std::cin>>x;
builds.push_back(x);
}
std::sort(builds.begin(),builds.end());
if(n<=14){
std::vector<std::vector<std::vector<int>>> vec(1<<n,std::vector<std::vector<int>>(n,std::vector<int>(l+1)));
for(int i=0;i<n;i++){
vec[1<<i][i][0]=1;
}
for(int e=0;e<1<<n;e++){
if(e==(e&-e))continue;
for(int x=0;x<n;x++){
if(((e>>x)&1)==0)continue;
int pid=e^(1<<x);
std::vector<int> cur(l+1);
for(int y=0;y<n;y++){
if(((e>>y)&1)==0||x==y)continue;
int offs=builds[x]-builds[y];
if(offs<0)offs=-offs;
for(int c=offs;c<=l;c++){
cur[c]+=vec[pid][y][c-offs];
cur[c]%=1000000007;
}
}
vec[e][x]=cur;
}
}
int result=0;
for(int i=0;i<n;i++){
for(int j=0;j<=l;j++){
result+=vec[(1<<n)-1][i][j];
result%=1000000007;
}
}
std::cout<<result<<'\n';
return 0;
}
}
