#include <bits/stdc++.h>
using namespace std;
int mod = 1e9 + 7;
int dp[1100007];
int main() {
  int n, k;
  long long int l;
  scanf("%d%I64d%d", &n, &l, &k);
  vector<int>(v);
  for (int i = (0); i < ((n)); ++i) {
    int(a);
    scanf("%d", &(a));
    ;
    (v).push_back(a);
  };
  vector<pair<int, int> > w;
  for (int i = (0); i < (n); ++i) w.push_back(make_pair(v[i], i));
  sort((w).begin(), (w).end());
  for (int i = (0); i < (n); ++i) dp[0 * n + i] = 1;
  for (int i = (1); i < (k); ++i) {
    int suma = 0;
    int lastj = 0;
    int j = 0;
    while (j != n) {
      while (j < n && w[j].first == w[lastj].first) {
        suma += dp[(i - 1) * n + w[j].second];
        suma %= mod;
        j++;
      }
      for (int u = (lastj); u < (j); ++u) {
        dp[i * n + w[u].second] = suma;
      }
      lastj = j;
    }
  }
  long long int pelne = l / n, reszta = l % n;
  int answer = 0;
  for (int i = (0); i < (k); ++i) {
    long long int mozl = max(pelne - i, 0LL);
    int dupa = 0;
    int dupa2 = 0;
    for (int j = (0); j < (n); ++j) {
      dupa += dp[i * n + j];
      dupa %= mod;
      if (j < reszta) {
        dupa2 += dp[i * n + j];
        dupa2 %= mod;
      }
    }
    mozl = mozl % mod;
    answer += (mozl * 1LL * dupa) % mod;
    answer %= mod;
    if (pelne >= i) {
      answer += dupa2;
      answer %= mod;
    }
  }
  printf("%d\n", answer);
}

### Prompt

Please create a solution in Cpp 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 mod = 1e9 + 7;
int dp[1100007];
int main() {
  int n, k;
  long long int l;
  scanf("%d%I64d%d", &n, &l, &k);
  vector<int>(v);
  for (int i = (0); i < ((n)); ++i) {
    int(a);
    scanf("%d", &(a));
    ;
    (v).push_back(a);
  };
  vector<pair<int, int> > w;
  for (int i = (0); i < (n); ++i) w.push_back(make_pair(v[i], i));
  sort((w).begin(), (w).end());
  for (int i = (0); i < (n); ++i) dp[0 * n + i] = 1;
  for (int i = (1); i < (k); ++i) {
    int suma = 0;
    int lastj = 0;
    int j = 0;
    while (j != n) {
      while (j < n && w[j].first == w[lastj].first) {
        suma += dp[(i - 1) * n + w[j].second];
        suma %= mod;
        j++;
      }
      for (int u = (lastj); u < (j); ++u) {
        dp[i * n + w[u].second] = suma;
      }
      lastj = j;
    }
  }
  long long int pelne = l / n, reszta = l % n;
  int answer = 0;
  for (int i = (0); i < (k); ++i) {
    long long int mozl = max(pelne - i, 0LL);
    int dupa = 0;
    int dupa2 = 0;
    for (int j = (0); j < (n); ++j) {
      dupa += dp[i * n + j];
      dupa %= mod;
      if (j < reszta) {
        dupa2 += dp[i * n + j];
        dupa2 %= mod;
      }
    }
    mozl = mozl % mod;
    answer += (mozl * 1LL * dupa) % mod;
    answer %= mod;
    if (pelne >= i) {
      answer += dupa2;
      answer %= mod;
    }
  }
  printf("%d\n", answer);
}

```