#include <bits/stdc++.h>
using namespace std;
const double PI = 3.141592653589793238;
bool DBG = false;
template <typename T, typename S>
std::ostream& operator<<(std::ostream& os, const std::pair<T, S>& p) {
  os << "(" << p.first << ", " << p.second << ")";
  return os;
}
template <typename T>
std::ostream& operator<<(std::ostream& os, const std::vector<T>& vector) {
  for (size_t i = 0; i < vector.size(); ++i) {
    os << vector[i] << " ";
  }
  return os;
}
template <typename T>
std::ostream& operator<<(std::ostream& os,
                         const std::vector<std::vector<T>>& vv) {
  for (size_t i = 0; i < vv.size(); ++i) {
    os << vv[i] << std::endl;
  }
  return os;
}
const int MOD = 1000000007;
int main() {
  std::ios_base::sync_with_stdio(false);
  int n, k;
  long long l;
  cin >> n >> l >> k;
  vector<pair<int, int>> a(n);
  for (int i = 0; i < (int)(n); ++i) {
    cin >> a[i].first;
    a[i].second = i;
  }
  if (DBG) {
    cout << "a init: \n" << a << endl;
  }
  sort(a.begin(), a.end(),
       [](const pair<int, int>& x, const pair<int, int>& y) -> bool {
         return x.first < y.first;
       });
  if (DBG) {
    cout << "a sort: \n" << a << endl;
  }
  vector<vector<int>> d(k + 1, vector<int>(n, 0));
  for (int i = 0; i < (int)(n); ++i) {
    d[1][i] = 1;
  }
  for (int len = 2; len <= k; ++len) {
    int pos = 0, sum = 0;
    for (int i = 0; i < (int)(n); ++i) {
      while (pos < n && a[pos].first <= a[i].first) {
        sum = (sum + d[len - 1][pos]) % MOD;
        ++pos;
      }
      d[len][i] = sum;
    }
  }
  long long full_blocks = l / n;
  int last_block_len = l % n;
  int res = 0;
  for (int len = 1; len <= k; ++len) {
    for (int i = 0; i < (int)(n); ++i) {
      int blocks_cnt = (full_blocks - len + 1) % MOD;
      if (a[i].second < last_block_len) {
        blocks_cnt++;
      }
      blocks_cnt = max(blocks_cnt, 0);
      long long add = 1ll * d[len][i] * blocks_cnt;
      res = (res + add % MOD) % MOD;
    }
  }
  cout << res << endl;
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

While Duff was resting in the beach, she accidentally found a strange array b0, b1, ..., bl - 1 consisting of l positive integers. This array was strange because it was extremely long, but there was another (maybe shorter) array, a0, ..., an - 1 that b can be build from a with formula: bi = ai mod n where a mod b denoted the remainder of dividing a by b.

<image>

Duff is so curious, she wants to know the number of subsequences of b like bi1, bi2, ..., bix (0 ≤ i1 < i2 < ... < ix < l), such that: 

  * 1 ≤ x ≤ k
  * For each 1 ≤ j ≤ x - 1, <image>
  * For each 1 ≤ j ≤ x - 1, bij ≤ bij + 1. i.e this subsequence is non-decreasing. 



Since this number can be very large, she want to know it modulo 109 + 7.

Duff is not a programmer, and Malek is unavailable at the moment. So she asked for your help. Please tell her this number.

Input

The first line of input contains three integers, n, l and k (1 ≤ n, k, n × k ≤ 106 and 1 ≤ l ≤ 1018).

The second line contains n space separated integers, a0, a1, ..., an - 1 (1 ≤ ai ≤ 109 for each 0 ≤ i ≤ n - 1). 

Output

Print the answer modulo 1 000 000 007 in one line.

Examples

Input

3 5 3
5 9 1


Output

10


Input

5 10 3
1 2 3 4 5


Output

25

Note

In the first sample case, <image>. So all such sequences are: <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image> and <image>.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const double PI = 3.141592653589793238;
bool DBG = false;
template <typename T, typename S>
std::ostream& operator<<(std::ostream& os, const std::pair<T, S>& p) {
  os << "(" << p.first << ", " << p.second << ")";
  return os;
}
template <typename T>
std::ostream& operator<<(std::ostream& os, const std::vector<T>& vector) {
  for (size_t i = 0; i < vector.size(); ++i) {
    os << vector[i] << " ";
  }
  return os;
}
template <typename T>
std::ostream& operator<<(std::ostream& os,
                         const std::vector<std::vector<T>>& vv) {
  for (size_t i = 0; i < vv.size(); ++i) {
    os << vv[i] << std::endl;
  }
  return os;
}
const int MOD = 1000000007;
int main() {
  std::ios_base::sync_with_stdio(false);
  int n, k;
  long long l;
  cin >> n >> l >> k;
  vector<pair<int, int>> a(n);
  for (int i = 0; i < (int)(n); ++i) {
    cin >> a[i].first;
    a[i].second = i;
  }
  if (DBG) {
    cout << "a init: \n" << a << endl;
  }
  sort(a.begin(), a.end(),
       [](const pair<int, int>& x, const pair<int, int>& y) -> bool {
         return x.first < y.first;
       });
  if (DBG) {
    cout << "a sort: \n" << a << endl;
  }
  vector<vector<int>> d(k + 1, vector<int>(n, 0));
  for (int i = 0; i < (int)(n); ++i) {
    d[1][i] = 1;
  }
  for (int len = 2; len <= k; ++len) {
    int pos = 0, sum = 0;
    for (int i = 0; i < (int)(n); ++i) {
      while (pos < n && a[pos].first <= a[i].first) {
        sum = (sum + d[len - 1][pos]) % MOD;
        ++pos;
      }
      d[len][i] = sum;
    }
  }
  long long full_blocks = l / n;
  int last_block_len = l % n;
  int res = 0;
  for (int len = 1; len <= k; ++len) {
    for (int i = 0; i < (int)(n); ++i) {
      int blocks_cnt = (full_blocks - len + 1) % MOD;
      if (a[i].second < last_block_len) {
        blocks_cnt++;
      }
      blocks_cnt = max(blocks_cnt, 0);
      long long add = 1ll * d[len][i] * blocks_cnt;
      res = (res + add % MOD) % MOD;
    }
  }
  cout << res << endl;
  return 0;
}

```