#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
long long l, ans;
map<int, int> id;
long long sum[1010000], tank[1010000];
int A[1010000], n, k, tp;
int main() {
  scanf("%d%I64d%d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &A[i]);
    id[A[i]] = 1;
  }
  tp = 0;
  for (map<int, int>::iterator i = id.begin(); i != id.end(); i++)
    i->second = ++tp;
  for (int i = 1; i <= tp; i++) sum[i] = 1;
  for (int i = 0; i < n; i++) A[i] = id[A[i]];
  for (int x = 1; x <= k; x++) {
    for (int i = 0; i < n; i++) {
      tank[i] = sum[A[i]];
      if (i < (l % n)) {
        if (((l / (long long)n) % mod + 1LL - (long long)x + 1LL) > 0)
          ans += sum[A[i]] *
                 ((l / (long long)n) % mod + 1LL - (long long)x + 1LL) % mod;
      } else {
        if ((l / (long long)n - (long long)x + 1LL) > 0)
          ans +=
              sum[A[i]] * ((l / (long long)n - (long long)x + 1LL) % mod) % mod;
      }
      ans %= mod;
    }
    for (int i = 1; i <= tp; i++) sum[i] = 0;
    for (int i = 0; i < n; i++) sum[A[i]] += tank[i], sum[A[i]] %= mod;
    for (int i = 1; i <= tp; i++) sum[i] += sum[i - 1], sum[i] %= mod;
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Your task is to create 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 = 1000000007;
long long l, ans;
map<int, int> id;
long long sum[1010000], tank[1010000];
int A[1010000], n, k, tp;
int main() {
  scanf("%d%I64d%d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &A[i]);
    id[A[i]] = 1;
  }
  tp = 0;
  for (map<int, int>::iterator i = id.begin(); i != id.end(); i++)
    i->second = ++tp;
  for (int i = 1; i <= tp; i++) sum[i] = 1;
  for (int i = 0; i < n; i++) A[i] = id[A[i]];
  for (int x = 1; x <= k; x++) {
    for (int i = 0; i < n; i++) {
      tank[i] = sum[A[i]];
      if (i < (l % n)) {
        if (((l / (long long)n) % mod + 1LL - (long long)x + 1LL) > 0)
          ans += sum[A[i]] *
                 ((l / (long long)n) % mod + 1LL - (long long)x + 1LL) % mod;
      } else {
        if ((l / (long long)n - (long long)x + 1LL) > 0)
          ans +=
              sum[A[i]] * ((l / (long long)n - (long long)x + 1LL) % mod) % mod;
      }
      ans %= mod;
    }
    for (int i = 1; i <= tp; i++) sum[i] = 0;
    for (int i = 0; i < n; i++) sum[A[i]] += tank[i], sum[A[i]] %= mod;
    for (int i = 1; i <= tp; i++) sum[i] += sum[i - 1], sum[i] %= mod;
  }
  printf("%I64d\n", ans);
  return 0;
}

```