#include <bits/stdc++.h>
using namespace std;
const int INF = (int)1e9;
const double EPS = (long double)1e-7;
const unsigned long long int mod = 1000000007;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  unsigned long long int n, l, k;
  cin >> n >> l >> k;
  map<unsigned long long int, unsigned long long int> a, al;
  for (unsigned long long int i = (0); i < (unsigned long long int)(n); ++i) {
    unsigned long long int x;
    cin >> x;
    a[x]++;
    if (i < l % n) al[x]++;
  }
  unsigned long long int ans = 0;
  unsigned long long int K = min(k, l / n);
  vector<vector<unsigned long long int> > dp(
      K + 10, vector<unsigned long long int>(a.size() + 10, 0));
  dp[0][0] = 1;
  for (unsigned long long int i = (1); i < (unsigned long long int)(K + 1);
       ++i) {
    unsigned long long int j = 0;
    unsigned long long int sm = 0;
    for (auto p : a) {
      sm = (sm + dp[i - 1][j]) % mod;
      dp[i][j] = (sm * p.second) % mod;
      ans += (((l / n - i + 1) % mod) * dp[i][j]) % mod;
      ans %= mod;
      j++;
    }
  }
  K = min(k, (l + n - 1) / n);
  for (unsigned long long int i = (1); i < (unsigned long long int)(K + 1);
       ++i) {
    unsigned long long int j = 0;
    unsigned long long int sm = 0;
    for (auto p : a) {
      sm = (sm + dp[i - 1][j]) % mod;
      if (al.count(p.first)) {
        ans += (sm * al[p.first]) % mod;
        ans %= mod;
      }
      j++;
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 INF = (int)1e9;
const double EPS = (long double)1e-7;
const unsigned long long int mod = 1000000007;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  unsigned long long int n, l, k;
  cin >> n >> l >> k;
  map<unsigned long long int, unsigned long long int> a, al;
  for (unsigned long long int i = (0); i < (unsigned long long int)(n); ++i) {
    unsigned long long int x;
    cin >> x;
    a[x]++;
    if (i < l % n) al[x]++;
  }
  unsigned long long int ans = 0;
  unsigned long long int K = min(k, l / n);
  vector<vector<unsigned long long int> > dp(
      K + 10, vector<unsigned long long int>(a.size() + 10, 0));
  dp[0][0] = 1;
  for (unsigned long long int i = (1); i < (unsigned long long int)(K + 1);
       ++i) {
    unsigned long long int j = 0;
    unsigned long long int sm = 0;
    for (auto p : a) {
      sm = (sm + dp[i - 1][j]) % mod;
      dp[i][j] = (sm * p.second) % mod;
      ans += (((l / n - i + 1) % mod) * dp[i][j]) % mod;
      ans %= mod;
      j++;
    }
  }
  K = min(k, (l + n - 1) / n);
  for (unsigned long long int i = (1); i < (unsigned long long int)(K + 1);
       ++i) {
    unsigned long long int j = 0;
    unsigned long long int sm = 0;
    for (auto p : a) {
      sm = (sm + dp[i - 1][j]) % mod;
      if (al.count(p.first)) {
        ans += (sm * al[p.first]) % mod;
        ans %= mod;
      }
      j++;
    }
  }
  cout << ans << endl;
  return 0;
}

```