Problem: 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.

Solution (in JAVA):

Solution:
/*
 * @author derrick20
 */
import java.io.*;
import java.util.*;

public class table {
    public static void main(String args[]) throws Exception {
        FastScanner sc = new FastScanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        int N = sc.nextInt();
        long M = sc.nextLong();
        int K = sc.nextInt();

        // choose[i][j] Represents i choose j
        long[][] choose = new long[N + 1][N + 1];
        for (int i = 0; i <= N; i++) {
            choose[i][0] = 1;
            choose[i][i] = 1;
            for (int j = 1; j < i; j++) {
                choose[i][j] = (choose[i - 1][j - 1] + choose[i - 1][j]) % mod;
            }
        }

        // dp[i][j] stores a solution bag of using up to and including the ith
        // column the number of ways to place j dots among them.
        long[][] dp = new long[N][K + 1];

        // aux[i][j] = N choose j, raised to the power appropriate depending on i and M % N
        long[][] aux = new long[N][N + 1];
        for (int i = 0; i < N; i++) {
            for (int j = 0; j <= N; j++) {
                // We want it so that if i < M % N, then it'll add 1 power.
                aux[i][j] = quicklyExponentiate(choose[N][j], M / N);
                if (i < M % N) {
                    aux[i][j] *= choose[N][j];
                    aux[i][j] %= mod;
                }
            }
        }

        // The key insight is that, during our transitions, we only need to look
        // at unique columns modulo N, since those with same remainder must have
        // the same number of dots, because the N - 1 columns between them are shared,
        // so by symmetry, they must match!
        // Base case: 1st column
        for (int i = 0; i < N; i++) {
            // Count how many times a column of a certain residue will occur
//            System.out.println(i);
            // Place all possible numbers of dots into this column residue class
            for (int dots = 0; dots <= K; dots++) {
                if (i == 0) {
                    // Base case: all dots go inside here
                    if (dots <= N) {
                        // With only the first column, we can only fit at most N dots
                        dp[i][dots] = aux[i][dots];
                    }
                }
                else {
                    for (int used = 0; used <= dots && used <= N; used++) {
                        // We can use up to N dots at most. The rest must be pushed off
                        // onto other columns
                        long myWays = aux[i][used];
                        // This is using the remaining dots to fill up the rest
                        // of the grid (using a solution bag already computed!)
                        long otherWays = dp[i - 1][dots - used];
                        dp[i][dots] += (myWays * otherWays) % mod;
                        dp[i][dots] %= mod;
                    }
                }
            }
        }
        // Given a total of K dots allowed to distribute, and the N column residues
        // available (up to the N-1 indexed one), we have all ways counted
        out.println(dp[N - 1][K]);
        out.close();
    }

    static long mod = (long) 1e9 + 7;

    static long quicklyExponentiate(long x, long p) {
        if (p == 0) {
            return 1;
        }
        else if (p == 1) {
            return x;
        }
        else {
            long sqrt = quicklyExponentiate(x, p / 2);
            long ans = (sqrt * sqrt) % mod;
            if (p % 2 == 0) {
                return ans;
            }
            else {
                return (x * ans) % mod;
            }
        }
    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        FastScanner(FileReader s) {
            br = new BufferedReader(s);
        }

        String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        String nextLine() throws IOException {
            return br.readLine();
        }

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

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

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