John Doe has an n × m table. John Doe can paint points in some table cells, not more than one point in one table cell. John Doe wants to use such operations to make each square subtable of size n × n have exactly k points.

John Doe wondered, how many distinct ways to fill the table with points are there, provided that the condition must hold. As this number can be rather large, John Doe asks to find its remainder after dividing by 1000000007 (109 + 7).

You should assume that John always paints a point exactly in the center of some cell. Two ways to fill a table are considered distinct, if there exists a table cell, that has a point in one way and doesn't have it in the other.

Input

A single line contains space-separated integers n, m, k (1 ≤ n ≤ 100; n ≤ m ≤ 1018; 0 ≤ k ≤ n2) — the number of rows of the table, the number of columns of the table and the number of points each square must contain.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. 

Output

In a single line print a single integer — the remainder from dividing the described number of ways by 1000000007 (109 + 7).

Examples

Input

5 6 1


Output

45

Note

Let's consider the first test case: 

<image> The gray area belongs to both 5 × 5 squares. So, if it has one point, then there shouldn't be points in any other place. If one of the white areas has a point, then the other one also must have a point. Thus, there are about 20 variants, where the point lies in the gray area and 25 variants, where each of the white areas contains a point. Overall there are 45 variants.
import java.io.*;
import static java.lang.Math.*;
import java.util.*;
import java.util.function.*;
import java.lang.*;

public class Main {
    final static boolean debug = false;
    final static String fileName = "";
    final static boolean useFiles = false;

    public static void main(String[] args) throws FileNotFoundException {
        long start;
        if (debug)
            System.nanoTime();
        InputStream inputStream;
        OutputStream outputStream;
        if (useFiles) {
            inputStream = new FileInputStream(fileName + ".in");
            outputStream = new FileOutputStream(fileName + ".out");
        } else {
            inputStream = System.in;
            outputStream = System.out;
        }
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task(in, out);
        solver.solve();
        if(debug)
            out.println((System.nanoTime() - start) / 1e+9);
        out.close();
    }
}
class Task {
    final int mod = (int) (1e+9 + 7);
    int[][] r;

    int power(int base, long pow, int mod) {
        long s = base;
        long result = 1;
        while (pow != 0) {
            if ((pow & 1) > 0) {
                result *= s;
                result %= mod;
            }
            s = (s * s) % mod;
            pow >>= 1;
        }
        return (int) result;
    }

    void initC(int n, long m){
        int[][] c = new int[n + 1][n + 1];
        for (int i = 0; i <= n; i++){
            c[i][0] = 1;
            for (int j = 1; j <= i; j++){
                c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
            }
        }
        r = new int[n][n + 1];
        for (int i = 0; i < n; i++) {
            for (int k = 0; k <= n; k++) {
                r[i][k] = power(c[n][k], (m - i + n - 1) / n, mod);
            }
        }
    }

    public void solve() {
        int n = in.nextInt();
        long m = in.nextLong();
        int k = in.nextInt();

        initC(n, m);

        long[][] d = new long[n][k + 1];
        for (int i = 0; i < n; i++){
            d[i][0] = 1;
        }
        for (int count = 0; count <= min(n, k); count++){
            d[0][count] = r[0][count];
        }
        for (int i = 1; i < n; i++) {
            for (int sum = 1; sum <= min((i + 1) * n, k); sum++) {
                for (int count = 0; count <= n && count <= sum; count++) {
                    d[i][sum] += d[i - 1][sum - count] * r[i][count];
                    d[i][sum] %= mod;
                }
            }
        }
        out.println(d[n - 1][k]);
    }

    private InputReader in;
    private PrintWriter out;

    Task(InputReader in, PrintWriter out) {
        this.in = in;
        this.out = out;
    }
}

class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream), 32768);
        tokenizer = null;
    }

    public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }

    public double nextDouble(){
        return Double.parseDouble(next());
    }

    public int nextInt() {
        return Integer.parseInt(next());
    }

    public long nextLong(){
        return Long.parseLong(next());
    }

    public byte nextByte(){
        return Byte.parseByte(next());
    }
}