#include <bits/stdc++.h>

using namespace std;



long nCrModpDP(long n, long r, long p)
{
    // The array C is going to store last row of
    // pascal triangle at the end. And last entry
    // of last row is nCr
    long C[r+1];
    memset(C, 0, sizeof(C));
 
    C[0] = 1; // Top row of Pascal Triangle
 
    // One by constructs remaining rows of Pascal
    // Triangle from top to bottom
    for (long i = 1; i <= n; i++)
    {
        // Fill entries of current row using previous
        // row values
        for (long j = min(i, r); j > 0; j--)
 
            // nCj = (n-1)Cj + (n-1)C(j-1);
            C[j] = (C[j] + C[j-1])%p;
    }
    return C[r];
}
 
// Lucas Theorem based function that returns nCr % p
// This function works like decimal to binary conversion
// recursive function.  First we compute last digits of
// n and r in base p, then recur for remaining digits
long nCrModpLucas(long n, long r, long p)
{
   // Base case
   if (r==0)
      return 1;
 
   // Compute last digits of n and r in base p
   long ni = n%p, ri = r%p;
 
   // Compute result for last digits computed above, and
   // for remaining digits.  Multiply the two results and
   // compute the result of multiplication in modulo p.
   return (nCrModpLucas(n/p, r/p, p) * // Last digits of n and r
           nCrModpDP(ni, ri, p)) % p;  // Remaining digits
}

long countArray(int n, int k, int x) {
    // Return the number of ways to fill in the array.
    
    return nCrModpLucas(x+1, n-2, 100000007);
}

int main() {
    int n;
    int k;
    int x;
    cin >> n >> k >> x;
    long answer = countArray(n, k, x);
    cout << answer << endl;
    return 0;
}