#include <bits/stdc++.h>
using namespace std;
int n;
long long l;
int k;
int a[1000005];
int b[1000005];
int len;
int c[1000005];
long long way[1000005];
const int Mod = 1000000007;
int main() {
  cin >> n >> l >> k;
  for (int i = (int)(0); i < (int)(n); ++i) scanf("%d", a + i), b[i] = a[i];
  sort(b, b + n);
  auto pt = unique(b, b + n);
  len = pt - b;
  for (int i = (int)(0); i < (int)(n); ++i) a[i] = lower_bound(b, pt, a[i]) - b;
  for (int i = (int)(0); i < (int)(n); ++i) ++c[a[i]];
  for (int i = (int)(0); i < (int)(n); ++i) way[i] = 1;
  long long R = 0;
  for (int ttt = (int)(0); ttt < (int)(k); ++ttt) {
    for (long long i = (long long)(max((long long)0, l - n));
         i < (long long)(l); ++i) {
      if (i / n < ttt) continue;
      R += way[a[i % n]] * ((i / n - ttt + 1) % Mod);
      R %= Mod;
    }
    for (int i = (int)(0); i < (int)(len); ++i) way[i] = (way[i] * c[i]) % Mod;
    for (int i = (int)(1); i < (int)(len); ++i) {
      way[i] += way[i - 1];
      if (way[i] >= Mod) way[i] -= Mod;
    }
  }
  cout << R << endl;
  return 0;
}

### Prompt

Generate 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;
int n;
long long l;
int k;
int a[1000005];
int b[1000005];
int len;
int c[1000005];
long long way[1000005];
const int Mod = 1000000007;
int main() {
  cin >> n >> l >> k;
  for (int i = (int)(0); i < (int)(n); ++i) scanf("%d", a + i), b[i] = a[i];
  sort(b, b + n);
  auto pt = unique(b, b + n);
  len = pt - b;
  for (int i = (int)(0); i < (int)(n); ++i) a[i] = lower_bound(b, pt, a[i]) - b;
  for (int i = (int)(0); i < (int)(n); ++i) ++c[a[i]];
  for (int i = (int)(0); i < (int)(n); ++i) way[i] = 1;
  long long R = 0;
  for (int ttt = (int)(0); ttt < (int)(k); ++ttt) {
    for (long long i = (long long)(max((long long)0, l - n));
         i < (long long)(l); ++i) {
      if (i / n < ttt) continue;
      R += way[a[i % n]] * ((i / n - ttt + 1) % Mod);
      R %= Mod;
    }
    for (int i = (int)(0); i < (int)(len); ++i) way[i] = (way[i] * c[i]) % Mod;
    for (int i = (int)(1); i < (int)(len); ++i) {
      way[i] += way[i - 1];
      if (way[i] >= Mod) way[i] -= Mod;
    }
  }
  cout << R << endl;
  return 0;
}

```