Problem: Anya loves to fold and stick. Today she decided to do just that.

Anya has n cubes lying in a line and numbered from 1 to n from left to right, with natural numbers written on them. She also has k stickers with exclamation marks. We know that the number of stickers does not exceed the number of cubes.

Anya can stick an exclamation mark on the cube and get the factorial of the number written on the cube. For example, if a cube reads 5, then after the sticking it reads 5!, which equals 120.

You need to help Anya count how many ways there are to choose some of the cubes and stick on some of the chosen cubes at most k exclamation marks so that the sum of the numbers written on the chosen cubes after the sticking becomes equal to S. Anya can stick at most one exclamation mark on each cube. Can you do it?

Two ways are considered the same if they have the same set of chosen cubes and the same set of cubes with exclamation marks.

Input

The first line of the input contains three space-separated integers n, k and S (1 ≤ n ≤ 25, 0 ≤ k ≤ n, 1 ≤ S ≤ 1016) — the number of cubes and the number of stickers that Anya has, and the sum that she needs to get. 

The second line contains n positive integers ai (1 ≤ ai ≤ 109) — the numbers, written on the cubes. The cubes in the input are described in the order from left to right, starting from the first one. 

Multiple cubes can contain the same numbers.

Output

Output the number of ways to choose some number of cubes and stick exclamation marks on some of them so that the sum of the numbers became equal to the given number S.

Examples

Input

2 2 30
4 3


Output

1


Input

2 2 7
4 3


Output

1


Input

3 1 1
1 1 1


Output

6

Note

In the first sample the only way is to choose both cubes and stick an exclamation mark on each of them.

In the second sample the only way is to choose both cubes but don't stick an exclamation mark on any of them.

In the third sample it is possible to choose any of the cubes in three ways, and also we may choose to stick or not to stick the exclamation mark on it. So, the total number of ways is six.

Solution (in JAVA):

Solution:
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	[_] ⠀⠀⠀⠀⠀[___]                     */
// Main Code at the Bottom
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main {
	//Fast IO class
    static class FastReader {
        BufferedReader br; 
        StringTokenizer st; 
        public FastReader() {
        	boolean env=System.getProperty("ONLINE_JUDGE") != null;
        	if(!env) {
        		try {
					br=new BufferedReader(new FileReader("src\\input.txt"));
				} catch (FileNotFoundException e) {
					e.printStackTrace();
				}
        	}
        	else br = new BufferedReader(new InputStreamReader(System.in)); 
        } 
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) {
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
        int nextInt() {
            return Integer.parseInt(next()); 
        } 
        long nextLong() {
            return Long.parseLong(next()); 
        } 
        double nextDouble() {
            return Double.parseDouble(next()); 
        } 
        String nextLine() {
            String str = ""; 
            try {
                str = br.readLine(); 
            } 
            catch (IOException e) {
                e.printStackTrace(); 
            } 
            return str; 
        } 
    }     
    static long MOD=1000000000+7;
    //Euclidean Algorithm
    static long gcd(long A,long B){
        if(B==0) return A;
        return gcd(B,A%B);
    }
    //Modular Exponentiation
    static long fastExpo(long x,long n){
        if(n==0) return 1;
        if((n&1)==0) return fastExpo((x*x)%MOD,n/2)%MOD;
        return ((x%MOD)*fastExpo((x*x)%MOD,(n-1)/2))%MOD;
    }
    //Modular Inverse
    static long inverse(long x) {
    	return fastExpo(x,MOD-2);
    }
    //Prime Number Algorithm
    static boolean isPrime(long n){
        if(n<=1) return false;
        if(n<=3) return true;
        if(n%2==0 || n%3==0) return false;
        for(int i=5;i*i<=n;i+=6) if(n%i==0 || n%(i+2)==0) return false;
        return true;
    }
    //Reverse an array
    static void reverse(int arr[],int l,int r){
    	while(l<r) {
    		int tmp=arr[l];
    		arr[l++]=arr[r];
    		arr[r++]=tmp;
    	}
    }
    //Print array
    static void print1d(int arr[]) {
    	out.println(Arrays.toString(arr));
    }
    static void print2d(int arr[][]) {
    	for(int a[]: arr) out.println(Arrays.toString(a));
    }
    // Pair
    static class pair{
    	long x,y;
    	pair(long a,long b){
    		this.x=a;
    		this.y=b;
    	}
    	public boolean equals(Object obj) {
    		if(obj == null || obj.getClass()!= this.getClass()) return false;
            pair p = (pair) obj;
            return (this.x==p.x && this.y==p.y);
        }
    	public int hashCode() {
            return Objects.hash(x,y);
        }
    }
    static FastReader sc=new FastReader();
    static PrintWriter out=new PrintWriter(System.out);
    //Main function(The main code starts from here)
    public static void main (String[] args) throws java.lang.Exception {
    	int test=1;
        while(test-->0) {
        	long fact[]=new long[19];
        	fact[0]=1L;
        	for(int i=1;i<19;i++) fact[i]=i*fact[i-1];
        	int n=sc.nextInt(),k=sc.nextInt();
        	long s=sc.nextLong(),a[]=new long[n],ans=0L;
        	for(int i=0;i<n;i++) a[i]=sc.nextLong();
        	HashMap<Long,int []> map=new HashMap<>();
        	for(int i=0;i<(1<<n/2);i++) {
        		int len=Integer.bitCount(i),idx=0;
        		long tmp[]=new long[len];
        		for(int j=0;j<n/2;j++) if((1&(i>>j))==1) tmp[idx++]=a[j];
        		for(int x=0;x<(1<<len);x++) {
        			int exc=Integer.bitCount(x);
        			long sum=0L,flag=0;
        			for(int j=0;j<len;j++) {
        				if((1&(x>>j))==1) {
        					if(tmp[j]>=19) {
        						flag=1;
        						continue;
        					}
        					sum+=fact[(int)tmp[j]];
        				}
        				else sum+=tmp[j];
        			}
        			if(flag==1 || exc>k || sum>s) continue;
        			if(map.get(sum)==null) map.put(sum, new int[26]);
        			int c[]=map.get(sum);
        			c[exc]++;
        			map.put(sum,c);
        		}
        	}
        	for(int i=0;i<(1<<(n-n/2));i++) {
        		int len=Integer.bitCount(i),idx=0;
        		long tmp[]=new long[len];
        		for(int j=n/2;j<n;j++) if((1&(i>>(j-n/2)))==1) tmp[idx++]=a[j];
        		for(int x=0;x<(1<<len);x++) {
        			int exc=Integer.bitCount(x);
        			long sum=0L,flag=0;
        			for(int j=0;j<len;j++) {
        				if((1&(x>>j))==1) {
        					if(tmp[j]>=19) {
        						flag=1;
        						continue;
        					}
        					sum+=fact[(int)tmp[j]];
        				}
        				else sum+=tmp[j];
        			}
        			if(flag==1 || exc>k || sum>s) continue;
        			if(map.get(s-sum)==null) continue;
        			int max_idx=k-exc;
        			int cnt[]=map.get(s-sum);
        			for(int c=max_idx;c>=0;c--) ans+=cnt[c];
        		}
        	}
        	out.println(ans);
        }
        out.flush();
        out.close();
    }
}