#include <bits/stdc++.h>
using namespace std;
long long min(long long a, long long b) { return a < b ? a : b; }
int main() {
  long long n, k, l;
  scanf("%lld%lld%lld", &n, &l, &k);
  long long a1[n], a2[n];
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a1[i]);
    a2[i] = a1[i];
  }
  sort(a1, a1 + n);
  long long dp[k + 1][n + 1];
  for (int i = 0; i <= k; i++)
    for (int j = 0; j <= n; j++) dp[i][j] = 0;
  for (int i = 0; i <= k; i++)
    for (int j = 1; j <= n; j++) {
      if (i == 0)
        dp[i][j] = 1;
      else {
        long long tmp3 = upper_bound(a1, a1 + n, a1[j - 1]) - a1;
        dp[i][j] = (dp[i][j - 1] % 1000000007 + dp[i - 1][tmp3] % 1000000007) %
                   1000000007;
      }
    }
  long long tmp1 = l / n, tmp2 = l % n, ans = 0;
  for (long long x = 1; x <= min(k, tmp1 + 1); x++) {
    ans = (ans + (dp[x][n] * ((tmp1 - x + 1) % 1000000007)) % 1000000007) %
          1000000007;
    for (long long j = 0; j < tmp2; j++) {
      long long tmp3 = upper_bound(a1, a1 + n, a2[j]) - a1;
      ans = (ans + dp[x - 1][tmp3]) % 1000000007;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
long long min(long long a, long long b) { return a < b ? a : b; }
int main() {
  long long n, k, l;
  scanf("%lld%lld%lld", &n, &l, &k);
  long long a1[n], a2[n];
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a1[i]);
    a2[i] = a1[i];
  }
  sort(a1, a1 + n);
  long long dp[k + 1][n + 1];
  for (int i = 0; i <= k; i++)
    for (int j = 0; j <= n; j++) dp[i][j] = 0;
  for (int i = 0; i <= k; i++)
    for (int j = 1; j <= n; j++) {
      if (i == 0)
        dp[i][j] = 1;
      else {
        long long tmp3 = upper_bound(a1, a1 + n, a1[j - 1]) - a1;
        dp[i][j] = (dp[i][j - 1] % 1000000007 + dp[i - 1][tmp3] % 1000000007) %
                   1000000007;
      }
    }
  long long tmp1 = l / n, tmp2 = l % n, ans = 0;
  for (long long x = 1; x <= min(k, tmp1 + 1); x++) {
    ans = (ans + (dp[x][n] * ((tmp1 - x + 1) % 1000000007)) % 1000000007) %
          1000000007;
    for (long long j = 0; j < tmp2; j++) {
      long long tmp3 = upper_bound(a1, a1 + n, a2[j]) - a1;
      ans = (ans + dp[x - 1][tmp3]) % 1000000007;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```