#include <bits/stdc++.h>
typedef long long LL;
using namespace std;
const int M = 1000000007;
int main() {
  ios_base::sync_with_stdio(0);
  cout.precision(15);
  cout.setf(ios::fixed);
  LL n, i, j, m, l, k;
  cin >> n >> l >> k;
  vector<LL> v(n), u(n);
  for (i = 0; i < n; i++) {
    cin >> v[i];
    u[i] = v[i];
  }
  LL l1 = l / n;
  LL l2 = l % n;
  LL dp[n][k + 1];
  LL ans = 0;
  for (i = 0; i < n; i++)
    for (j = 0; j <= k; j++) dp[i][j] = 0;
  sort(u.begin(), u.end());
  for (j = 1; j <= k; j++) {
    LL pt = 0;
    for (i = 0; i < n; i++) {
      while (pt < n && u[pt] <= u[i]) pt++;
      auto it = upper_bound(u.begin(), u.end(), u[i]);
      LL pos = distance(u.begin(), it);
      dp[i][j] =
          ((j > 1 ? dp[pos - 1][j - 1] : 1) + (i > 0 ? dp[i - 1][j] : 0)) % M;
      if (i == (n - 1))
        ans = (ans + (l1 >= j ? (dp[i][j] * ((l1 - j + 1) % M)) : 0)) % M;
    }
  }
  for (i = 0; i < l2; i++)
    for (j = 1; j <= k; j++) {
      auto it = upper_bound(u.begin(), u.end(), v[i]);
      LL pos = distance(u.begin(), it);
      if ((j - 1) <= l1) ans = (ans + (j > 1 ? dp[pos - 1][j - 1] : 1)) % M;
    }
  cout << ans << '\n';
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
typedef long long LL;
using namespace std;
const int M = 1000000007;
int main() {
  ios_base::sync_with_stdio(0);
  cout.precision(15);
  cout.setf(ios::fixed);
  LL n, i, j, m, l, k;
  cin >> n >> l >> k;
  vector<LL> v(n), u(n);
  for (i = 0; i < n; i++) {
    cin >> v[i];
    u[i] = v[i];
  }
  LL l1 = l / n;
  LL l2 = l % n;
  LL dp[n][k + 1];
  LL ans = 0;
  for (i = 0; i < n; i++)
    for (j = 0; j <= k; j++) dp[i][j] = 0;
  sort(u.begin(), u.end());
  for (j = 1; j <= k; j++) {
    LL pt = 0;
    for (i = 0; i < n; i++) {
      while (pt < n && u[pt] <= u[i]) pt++;
      auto it = upper_bound(u.begin(), u.end(), u[i]);
      LL pos = distance(u.begin(), it);
      dp[i][j] =
          ((j > 1 ? dp[pos - 1][j - 1] : 1) + (i > 0 ? dp[i - 1][j] : 0)) % M;
      if (i == (n - 1))
        ans = (ans + (l1 >= j ? (dp[i][j] * ((l1 - j + 1) % M)) : 0)) % M;
    }
  }
  for (i = 0; i < l2; i++)
    for (j = 1; j <= k; j++) {
      auto it = upper_bound(u.begin(), u.end(), v[i]);
      LL pos = distance(u.begin(), it);
      if ((j - 1) <= l1) ans = (ans + (j > 1 ? dp[pos - 1][j - 1] : 1)) % M;
    }
  cout << ans << '\n';
}

```