This problem is same as the previous one, but has larger constraints.

Aki is playing a new video game. In the video game, he will control Neko, the giant cat, to fly between planets in the Catniverse.

There are n planets in the Catniverse, numbered from 1 to n. At the beginning of the game, Aki chooses the planet where Neko is initially located. Then Aki performs k - 1 moves, where in each move Neko is moved from the current planet x to some other planet y such that:

  * Planet y is not visited yet. 
  * 1 ≤ y ≤ x + m (where m is a fixed constant given in the input) 



This way, Neko will visit exactly k different planets. Two ways of visiting planets are called different if there is some index i such that, the i-th planet visited in the first way is different from the i-th planet visited in the second way.

What is the total number of ways to visit k planets this way? Since the answer can be quite large, print it modulo 10^9 + 7.

Input

The only line contains three integers n, k and m (1 ≤ n ≤ 10^9, 1 ≤ k ≤ min(n, 12), 1 ≤ m ≤ 4) — the number of planets in the Catniverse, the number of planets Neko needs to visit and the said constant m.

Output

Print exactly one integer — the number of different ways Neko can visit exactly k planets. Since the answer can be quite large, print it modulo 10^9 + 7.

Examples

Input


3 3 1


Output


4


Input


4 2 1


Output


9


Input


5 5 4


Output


120


Input


100 1 2


Output


100

Note

In the first example, there are 4 ways Neko can visit all the planets:

  * 1 → 2 → 3 
  * 2 → 3 → 1 
  * 3 → 1 → 2 
  * 3 → 2 → 1 



In the second example, there are 9 ways Neko can visit exactly 2 planets:

  * 1 → 2 
  * 2 → 1 
  * 2 → 3 
  * 3 → 1 
  * 3 → 2 
  * 3 → 4 
  * 4 → 1 
  * 4 → 2 
  * 4 → 3 



In the third example, with m = 4, Neko can visit all the planets in any order, so there are 5! = 120 ways Neko can visit all the planets.

In the fourth example, Neko only visit exactly 1 planet (which is also the planet he initially located), and there are 100 ways to choose the starting planet for Neko.
#include <bits/stdc++.h>
using namespace std;
const int maxN = 1e5 + 10, mod = 1e9 + 7, maxHav = 13, maxST = (1 << 5) + 1;
int dotN, jumpK, disM, colC;
void add(int &fir, long long sec) { fir = (long long)(fir + sec) % mod; }
struct Matrix {
  int num[210][210], row, col;
  void init(int sze) {
    row = col = sze;
    memset(num, 0, sizeof(num));
    for (int rowI = 0; rowI < row; ++rowI) {
      num[rowI][rowI] = 1;
    }
  }
  void init2(int r, int c) {
    row = r, col = c;
    memset(num, 0, sizeof(num));
  }
  Matrix operator=(const Matrix d) {
    row = d.row;
    col = d.col;
    for (int rowI = 0; rowI < row; ++rowI) {
      for (int colJ = 0; colJ < col; ++colJ) {
        num[rowI][colJ] = d.num[rowI][colJ];
      }
    }
    return (*this);
  }
  Matrix operator*(const Matrix d) const {
    Matrix ret;
    memset(ret.num, 0, sizeof(ret.num));
    ret.row = row;
    ret.col = d.col;
    for (int rowI = 0; rowI < ret.row; ++rowI) {
      for (int colJ = 0; colJ < ret.col; ++colJ) {
        for (int mid = 0; mid < col; ++mid) {
          add(ret.num[rowI][colJ], 1ll * num[rowI][mid] * d.num[mid][colJ]);
        }
      }
    }
    return ret;
  }
  Matrix power(int up) const {
    Matrix ret, base = (*this);
    ret.init(row);
    while (up) {
      if (up & 1) ret = ret * base;
      base = base * base;
      up >>= 1;
    }
    return ret;
  }
} ans, base, orig;
int pos(int er, int san) { return er * colC + san; }
int bitCount(int st) {
  int ret = 0;
  for (int weiI = 0; weiI < 4; ++weiI) {
    if (st & (1 << weiI)) {
      ++ret;
    }
  }
  return ret;
}
void init() {
  scanf("%d%d%d", &dotN, &jumpK, &disM);
  colC = 1 << disM;
  ans.init(colC * (jumpK + 1));
  orig.init2(1, colC * (jumpK + 1));
  orig.num[0][0] = 1;
  base.init2(colC * (jumpK + 1), colC * (jumpK + 1));
  for (int jumpJ = 0; jumpJ <= jumpK; ++jumpJ) {
    for (int stK = 0; stK < (1 << disM); ++stK) {
      if (!(stK & (1 << disM - 1))) {
        add(base.num[pos(jumpJ, stK << 1)][pos(jumpJ, stK)], 1);
        add(base.num[pos(jumpJ, stK << 1 | 1)][pos(jumpJ, stK)], 1);
      } else {
        if (jumpJ >= 1) {
          add(base.num[pos(jumpJ - 1, (stK - (1 << disM - 1)) << 1)]
                      [pos(jumpJ, stK)],
              1ll * (1 + bitCount((stK - (1 << disM - 1)) << 1)));
          add(base.num[pos(jumpJ - 1, (stK - (1 << disM - 1)) << 1 | 1)]
                      [pos(jumpJ, stK)],
              1ll * (1 + bitCount((stK - (1 << disM - 1)) << 1 | 1)));
        }
      }
    }
  }
}
int main() {
  init();
  int cnt = 0;
  ans = orig * base.power(dotN);
  for (int stK = 0; stK < (1 << disM); ++stK)
    add(cnt, ans.num[0][pos(jumpK, stK)]);
  printf("%d\n", cnt);
  return 0;
}
