#include <bits/stdc++.h>
using namespace std;
const int MAX = 1100000;
const int MOD = 1000000000 + 7;
int n, k, a[MAX], s[MAX];
long long cur[MAX], len;
int main() {
  cin >> n >> len >> k;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    s[i] = a[i];
  }
  sort(s, s + n);
  long long full = len / n;
  map<int, int> rem;
  for (int i = 0; i < len % n; i++) {
    ++rem[a[i]];
  }
  int ans = 0;
  cur[0] = 1;
  for (int i = 0; i < n; i++)
    if (i + 1 == n || s[i] < s[i + 1]) {
      int cnt = 1;
      while (i - cnt >= 0 && s[i - cnt] == s[i]) {
        ++cnt;
      }
      for (int j = 1; j <= k; j++) {
        cur[j] = (cur[j] + (long long)cur[j - 1] * cnt) % MOD;
      }
      int last = rem[s[i]];
      for (int j = 1; j <= k && full >= j - 1; j++) {
        ans = (ans + (long long)cur[j - 1] * last) % MOD;
      }
    }
  for (int j = 1; j <= k && full - j + 1 >= 0; j++) {
    ans = (ans + (long long)cur[j] * (full % MOD + MOD - j + 1)) % MOD;
  }
  printf("%d\n", ans);
  return 0;
}

### 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 MAX = 1100000;
const int MOD = 1000000000 + 7;
int n, k, a[MAX], s[MAX];
long long cur[MAX], len;
int main() {
  cin >> n >> len >> k;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    s[i] = a[i];
  }
  sort(s, s + n);
  long long full = len / n;
  map<int, int> rem;
  for (int i = 0; i < len % n; i++) {
    ++rem[a[i]];
  }
  int ans = 0;
  cur[0] = 1;
  for (int i = 0; i < n; i++)
    if (i + 1 == n || s[i] < s[i + 1]) {
      int cnt = 1;
      while (i - cnt >= 0 && s[i - cnt] == s[i]) {
        ++cnt;
      }
      for (int j = 1; j <= k; j++) {
        cur[j] = (cur[j] + (long long)cur[j - 1] * cnt) % MOD;
      }
      int last = rem[s[i]];
      for (int j = 1; j <= k && full >= j - 1; j++) {
        ans = (ans + (long long)cur[j - 1] * last) % MOD;
      }
    }
  for (int j = 1; j <= k && full - j + 1 >= 0; j++) {
    ans = (ans + (long long)cur[j] * (full % MOD + MOD - j + 1)) % MOD;
  }
  printf("%d\n", ans);
  return 0;
}

```