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#include <algorithm>
#include <iostream>
#include <vector>
#include <list>
#include <cassert>
#include <map>
#include <numeric>
using namespace std;

#define FOR(i, n) for(int i = 0, __n = (n); i < __n; i++)


static const long long MOD = 1000000007LL;

struct Frac {
  long long v;

  Frac(long long x = 0) {
      x %= MOD;
      if (x < 0) x += MOD;
      v = x;
  }

  Frac(long long num, long long den) {
      num %= MOD;
      if (num < 0) num += MOD;
      den %= MOD;
      if (den < 0) den += MOD;
      v = num * Frac(den).inv().v % MOD;
  }

  Frac pow(long long e) const {
    long long a = v;;
    long long r = 1;
    while (e > 0) {
        if (e & 1) r = r * a % MOD;
        a = a * a % MOD;
        e >>= 1;
    }
    return r;
  } 

  Frac inv() const {
      return pow(MOD - 2);
  }

  Frac operator+(const Frac& other) const {
      return Frac(v + other.v);
  }

  Frac operator-(const Frac& other) const {
      return Frac(v - other.v);
  }

  Frac operator*(const Frac& other) const {
      return Frac(v * other.v % MOD);
  }

  Frac operator/(const Frac& other) const {
      return Frac(v * other.inv().v % MOD);
  }

  Frac& operator+=(const Frac& other) {
      v += other.v;
      if (v >= MOD) v -= MOD;
      return *this;
  }

  Frac& operator-=(const Frac& other) {
      v -= other.v;
      if (v < 0) v += MOD;
      return *this;
  }

};

const int MAXN = 1000010;

Frac p[MAXN]; // prawdopodobienstwo dojscia 1 gracza do x w grze 1-os
Frac r[MAXN+1]; // prawdopodobienstwo, ze gra sie skończy od x (1os)

int main() {
  int n, k, m;
  cin >> n >> k >> m;

  Frac inv_k = Frac(1, k);

  int lim = max(0, m - k);

  p[0] = 1;
  Frac window = 0;
  FOR(i, m) {
    window += p[i] * inv_k;
    if (i >= k) {
      window -= p[i-k] * inv_k;
    }
    p[i+1] = window;
  }

  r[m] = 0;
  for(int i = m-1; i >= 0; i--) {
    if (i >= lim) {
      Frac koniec = p[i] * inv_k * Frac(i+k-m+1);
      r[i] = r[i+1] + koniec;
    } else {
      r[i] = 1;
    }
    
  }

  Frac res = 0;
  FOR(i, m) {
    if(i < lim) {
      res += p[i] * n;
    } else {
      res += p[i] * (r[i].pow(n) - r[i+1].pow(n))/(r[i] - r[i+1]);
    }
  }
  cout << res.v << endl;

  return 0;
}