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.
#include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &A) {
  for (long long i = 0; i < A.size(); i++) os << A[i] << " ";
  os << endl;
  return os;
}
template <>
ostream &operator<<(ostream &os, const vector<vector<long long> > &A) {
  long long N = A.size();
  for (long long i = 0; i < N; i++) {
    for (long long j = 0; j < A[i].size(); j++) os << A[i][j] << " ";
    os << endl;
  }
  return os;
}
struct edge {
  long long from, to, d, c;
  edge(long long _from = 0, long long _to = 0, long long _d = 0,
       long long _c = 0) {
    from = _from;
    to = _to;
    d = _d;
    c = _c;
  }
  bool operator<(const edge &rhs) const {
    return (d == rhs.d) ? (c < rhs.c) : (d < rhs.d);
  }
};
struct aabb {
  long long x1, y1, x2, y2;
  aabb(long long x1, long long y1, long long x2, long long y2)
      : x1(x1), y1(y1), x2(x2), y2(y2) {}
};
struct flow {
  long long to, cap, rev, cost;
  flow(long long to = 0, long long cap = 0, long long rev = 0,
       long long cost = 0)
      : to(to), cap(cap), rev(rev), cost(cost) {}
};
const long long di[4] = {0, -1, 0, 1};
const long long dj[4] = {-1, 0, 1, 0};
const long long ci[5] = {0, 0, -1, 0, 1};
const long long cj[5] = {0, -1, 0, 1, 0};
const long long LINF = LLONG_MAX / 2;
const long long INF = INT_MAX / 2;
const long double PI = acos(-1);
long long pow2(long long n) { return 1LL << n; }
template <typename T, typename U>
bool chmin(T &x, const U &y) {
  if (x > y) {
    x = y;
    return true;
  }
  return false;
}
template <typename T, typename U>
bool chmax(T &x, const U &y) {
  if (x < y) {
    x = y;
    return true;
  }
  return false;
}
struct initializer {
  initializer() { cout << fixed << setprecision(20); }
};
initializer _____;
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long extgcd(long long a, long long b, long long &x, long long &y) {
  long long d = a;
  if (b != 0) {
    d = extgcd(b, a % b, y, x);
    y -= (a / b) * x;
  } else {
    x = 1;
    y = 0;
  }
  return d;
}
long long factorial(long long x) {
  if (x == 1)
    return 1;
  else
    return x * factorial(x - 1);
}
long long Mfactorial(long long x, long long m) {
  long long ret = 1;
  for (long long i = 1; i <= x; i++) {
    ret *= i;
    ret %= m;
  }
  return ret;
}
vector<long long> MfactTable(long long n, long long m) {
  vector<long long> ret(n + 1);
  ret[0] = 1;
  long long tmp = 1;
  for (long long i = 1; i <= n; i++) {
    tmp *= i;
    tmp %= m;
    ret[i] = tmp;
  }
  return ret;
}
long long Minverse(long long a, long long m) {
  long long x, y;
  extgcd(a, m, x, y);
  return (m + x % m) % m;
}
long long McombSmall(long long n, long long k, long long m) {
  if (k > n) return 0;
  long long ret = Mfactorial(n, m);
  ret *= Minverse(Mfactorial(k, m), m);
  ret %= m;
  ret *= Minverse(Mfactorial(n - k, m), m);
  return ret % m;
}
long long McombFromFactTable(vector<long long> &T, long long n, long long k,
                             long long m) {
  if (k > n) return 0;
  long long ret = T[n];
  ret *= Minverse(T[k], m);
  ret %= m;
  ret *= Minverse(T[n - k], m);
  return ret % m;
}
long long pow(long long x, long long n) {
  if (n == 0) return 1;
  long long ret = (n & 1) ? x : 1;
  long long a = pow(x, n / 2);
  return ret * a * a;
}
long long Mpow(long long x, long long n, long long m) {
  if (n == 0) return 1;
  long long ret = (n & 1) ? x : 1;
  long long a = Mpow(x, n / 2, m);
  return (((ret * a) % m) * a) % m;
}
vector<vector<long long> > combTable(long long N) {
  vector<vector<long long> > ret(N + 1, vector<long long>(N + 1));
  for (long long i = 0; i <= N; i++)
    for (long long j = 0; j <= i; j++) {
      if (j == 0 || j == i)
        ret[i][j] = 1;
      else
        ret[i][j] = ret[i - 1][j - 1] + ret[i - 1][j];
    }
  return ret;
}
long long countNmodPisQlr(long long l, long long r, long long P, long long Q) {
  return (r - Q + P) / P - (l - Q + P - 1) / P;
}
struct Combination {
  long long N = 0, M;
  vector<long long> Ftable, FItable, Inversetable;
  void init() {
    Ftable.resize(N + 1);
    FItable.resize(N + 1);
    Inversetable.resize(N + 1);
    Ftable[0] = 1;
    FItable[0] = 1;
    for (long long i = 1; i <= N; i++) Ftable[i] = Ftable[i - 1] * i % M;
    FItable[N] = Minverse(Ftable[N], M);
    for (long long i = N - 1; i >= 1; i--)
      FItable[i] = FItable[i + 1] * (i + 1) % M;
    Inversetable[1] = 1;
    for (long long i = 2; i <= N; i++)
      Inversetable[i] = M - Inversetable[M % i] * (M / i) % M;
  }
  Combination(long long max, long long mod) {
    N = max;
    M = mod;
    init();
  }
  long long calc(long long n, long long k) {
    if (k > n) return 0;
    long long ret = Ftable[n] * FItable[k] % M * FItable[n - k] % M;
    return ret;
  }
  long long inverse(long long n) { return Inversetable[n]; }
  long long factorial(long long n) { return Ftable[n]; }
  long long factorialInverse(long long n) { return FItable[n]; }
  long long permutation(long long n, long long k) {
    if (k > n) return 0;
    return Ftable[n] * FItable[n - k] % M;
  }
};
long long N, M, K, T, Q, H, W;
signed main() {
  cin >> N >> M >> K;
  vector<vector<long long> > dp(N + 1, vector<long long>(K + 1, 0));
  dp[0][K] = 1;
  Combination cmb(N + 10, 1000000007);
  for (long long i = 0; i < (long long)(N); i++)
    for (long long k = 0; k < (long long)(N + 1); k++) {
      long long c = M / N;
      if (i < M % N) c++;
      long long tmp = Mpow(cmb.calc(N, k), c, 1000000007);
      for (long long j = 0; j < (long long)(K + 1); j++) {
        if (j - k < 0) continue;
        (dp[i + 1][j - k] += dp[i][j] * tmp % 1000000007) %= 1000000007;
      }
    }
  cout << dp[N][0] << endl;
  return 0;
}
