#include <bits/stdc++.h>
using namespace std;
const long long MOD = (long long)(1e9 + 7);
vector<int> tmp;
int main() {
  int n, k;
  long long l;
  cin >> n >> l >> k;
  vector<int> arr(n);
  for (int i = 0; i < (int)n; ++i) scanf("%d", &arr[i]);
  tmp = arr;
  sort(tmp.begin(), tmp.end());
  long long num = (l + n - 1) / n;
  k = (int)min((long long)k, num);
  vector<vector<int> > f(k + 1, vector<int>(n, 0));
  vector<vector<int> > prefix(k + 1, vector<int>(n, 0));
  for (int i = 0; i < (int)n; ++i) {
    f[1][i] = 1;
    prefix[1][i] = i + 1;
  }
  vector<int> aux(n);
  for (int i = 0; i < (int)n; ++i) {
    aux[i] = upper_bound(tmp.begin(), tmp.end(), arr[i]) - tmp.begin();
  }
  for (int i = 2; i <= k; ++i) {
    for (int j = 0; j < n; ++j) {
      int len = aux[j];
      f[i][j] = prefix[i - 1][len - 1];
      prefix[i][len - 1] += f[i][j];
      prefix[i][len - 1] %= MOD;
    }
    for (int j = 1; j < n; ++j) {
      prefix[i][j] = (prefix[i][j] + prefix[i][j - 1]) % MOD;
    }
  }
  int rem = l % n;
  long long res = 0;
  for (int i = 0; i < rem; ++i) {
    for (int len = 1; len <= k; ++len) {
      res += ((long long)f[len][i] * ((num - len + 1) % MOD)) % MOD;
      res %= MOD;
    }
  }
  if (rem == 0) ++num;
  for (int i = rem; i < n; ++i) {
    for (int len = 1; len <= k; ++len) {
      res += ((long long)f[len][i] * ((num - 1 - len + 1) % MOD)) % MOD;
      res %= 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 long long MOD = (long long)(1e9 + 7);
vector<int> tmp;
int main() {
  int n, k;
  long long l;
  cin >> n >> l >> k;
  vector<int> arr(n);
  for (int i = 0; i < (int)n; ++i) scanf("%d", &arr[i]);
  tmp = arr;
  sort(tmp.begin(), tmp.end());
  long long num = (l + n - 1) / n;
  k = (int)min((long long)k, num);
  vector<vector<int> > f(k + 1, vector<int>(n, 0));
  vector<vector<int> > prefix(k + 1, vector<int>(n, 0));
  for (int i = 0; i < (int)n; ++i) {
    f[1][i] = 1;
    prefix[1][i] = i + 1;
  }
  vector<int> aux(n);
  for (int i = 0; i < (int)n; ++i) {
    aux[i] = upper_bound(tmp.begin(), tmp.end(), arr[i]) - tmp.begin();
  }
  for (int i = 2; i <= k; ++i) {
    for (int j = 0; j < n; ++j) {
      int len = aux[j];
      f[i][j] = prefix[i - 1][len - 1];
      prefix[i][len - 1] += f[i][j];
      prefix[i][len - 1] %= MOD;
    }
    for (int j = 1; j < n; ++j) {
      prefix[i][j] = (prefix[i][j] + prefix[i][j - 1]) % MOD;
    }
  }
  int rem = l % n;
  long long res = 0;
  for (int i = 0; i < rem; ++i) {
    for (int len = 1; len <= k; ++len) {
      res += ((long long)f[len][i] * ((num - len + 1) % MOD)) % MOD;
      res %= MOD;
    }
  }
  if (rem == 0) ++num;
  for (int i = rem; i < n; ++i) {
    for (int len = 1; len <= k; ++len) {
      res += ((long long)f[len][i] * ((num - 1 - len + 1) % MOD)) % MOD;
      res %= MOD;
    }
  }
  cout << res << endl;
  return 0;
}

```