#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 10;
const long long P = 1e9 + 7;
pair<int, int> a[MAXN];
long long f[MAXN], N, K, L, r;
int main() {
  scanf("%d%I64d%d", &N, &L, &K);
  for (int i = 0; i < N; i++) {
    scanf("%d", &a[i].first);
    a[i].second = i;
    f[i] = 1;
  }
  sort(a, a + N);
  for (int n = N; n < N * K; n += N) {
    long long s = 0;
    for (int i = 0, j = 0; i < N; i = j) {
      j = i;
      while (j < N && a[i].first == a[j].first) {
        s = (s + f[n + j - N]) % P;
        j++;
      }
      for (int k = i; k < j; k++) f[n + k] = s;
    }
  }
  for (int i = 0; i < N * K; i++) {
    long long n = L / N + (a[i % N].second < L % N) - i / N;
    r = (r + max(0ll, n) % P * f[i]) % P;
  }
  printf("%I64d\n", r);
}

### 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 int MAXN = 1e6 + 10;
const long long P = 1e9 + 7;
pair<int, int> a[MAXN];
long long f[MAXN], N, K, L, r;
int main() {
  scanf("%d%I64d%d", &N, &L, &K);
  for (int i = 0; i < N; i++) {
    scanf("%d", &a[i].first);
    a[i].second = i;
    f[i] = 1;
  }
  sort(a, a + N);
  for (int n = N; n < N * K; n += N) {
    long long s = 0;
    for (int i = 0, j = 0; i < N; i = j) {
      j = i;
      while (j < N && a[i].first == a[j].first) {
        s = (s + f[n + j - N]) % P;
        j++;
      }
      for (int k = i; k < j; k++) f[n + k] = s;
    }
  }
  for (int i = 0; i < N * K; i++) {
    long long n = L / N + (a[i % N].second < L % N) - i / N;
    r = (r + max(0ll, n) % P * f[i]) % P;
  }
  printf("%I64d\n", r);
}

```