Problem: Yaroslav calls an array of r integers a1, a2, ..., ar good, if it meets the following conditions: |a1 - a2| = 1, |a2 - a3| = 1, ..., |ar - 1 - ar| = 1, |ar - a1| = 1, at that <image>. 

An array of integers b1, b2, ..., br is called great, if it meets the following conditions:

  1. The elements in it do not decrease (bi ≤ bi + 1). 
  2. If the inequalities 1 ≤ r ≤ n and 1 ≤ bi ≤ m hold. 
  3. If we can rearrange its elements and get at least one and at most k distinct good arrays. 



Yaroslav has three integers n, m, k. He needs to count the number of distinct great arrays. Help Yaroslav! As the answer may be rather large, print the remainder after dividing it by 1000000007 (109 + 7).

Two arrays are considered distinct if there is a position in which they have distinct numbers.

Input

The single line contains three integers n, m, k (1 ≤ n, m, k ≤ 100).

Output

In a single line print the remainder after dividing the answer to the problem by number 1000000007 (109 + 7).

Examples

Input

1 1 1


Output

0


Input

3 3 3


Output

2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
namespace mine {
long long qread() {
  long long ans = 0, f = 1;
  char c = getchar();
  while (c < '0' or c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while ('0' <= c and c <= '9') ans = ans * 10 + c - '0', c = getchar();
  return ans * f;
}
void write(long long num) {
  if (num < 0) putchar('-'), num = -num;
  if (num >= 10) write(num / 10);
  putchar('0' + num % 10);
}
void write1(long long num) {
  write(num);
  putchar(' ');
}
void write2(long long num) {
  write(num);
  putchar('\n');
}
template <typename T>
void chmax(T &x, const T y) {
  x = (x > y ? x : y);
}
template <typename T>
void chmin(T &x, const T y) {
  x = (x < y ? x : y);
}
long long gcd(long long x, long long y) { return y ? gcd(y, x % y) : 0; }
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
inline int mm(const int x) { return x >= MOD ? x - MOD : x; }
inline long long qpower(long long x, long long e) {
  long long ans = 1;
  while (e) {
    if (e & 1) ans = ans * x % MOD;
    x = x * x % MOD;
    e >>= 1;
  }
  return ans;
}
long long invm(long long x) { return qpower(x, MOD - 2); }
const int N = 1e2 + 10;
double C[N][N];
int f[2][N][N][N];
void main() {
  C[0][0] = 1;
  for (int i = 1; i < N; i++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++) C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
  }
  int n = qread(), m = qread(), k = qread();
  int ans = 0;
  for (int i = 1; i <= n; i++) f[0][i][i][1] = 1;
  for (int mx = 1, s = 0; mx <= m; mx++, s ^= 1) {
    memset(f[s ^ 1], 0, sizeof f[s ^ 1]);
    for (int now = 1; now <= n; now++)
      for (int cnt = 1; cnt <= k; cnt++)
        if (f[s][now][0][cnt])
          ans = (ans + (long long)f[s][now][0][cnt] * (m - mx + 1)) % MOD;
    for (int now = 1; now <= n; now++)
      for (int pp = 1; pp <= n; pp++)
        for (int T = pp; now + T <= n; T++)
          if (C[T - 1][pp - 1] <= k) {
            int tmp = C[T - 1][pp - 1];
            for (int cnt = 1; cnt * tmp <= k; cnt++)
              (f[s ^ 1][now + T][T - pp][cnt * tmp] += f[s][now][pp][cnt]) %=
                  MOD;
          }
  }
  write(ans);
}
};  // namespace mine
int main() {
  srand(time(0));
  mine::main();
}