#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
int a[1100000];
long long sm[2][1100000];
vector<int> vls;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  long long l;
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    vls.push_back(a[i]);
  }
  sort(vls.begin(), vls.end());
  vector<int>::iterator it;
  it = unique(vls.begin(), vls.end());
  vls.resize(distance(vls.begin(), it));
  for (int i = 0; i < n; i++)
    a[i] = (lower_bound(vls.begin(), vls.end(), a[i]) - vls.begin()) + 1;
  memset(sm, 0, sizeof sm);
  long long tot = 0;
  sm[1][0] = 1;
  sm[0][0] = 1;
  for (int j = 1; j <= n; j++) {
    sm[1][j] += sm[1][j - 1];
    sm[1][j] %= MOD;
    sm[0][j] += sm[1][j];
    sm[0][j] %= MOD;
  }
  for (int i = 0; i < k; i++) {
    for (int j = 0; j <= n; j++) sm[1][j] = 0;
    for (int j = 0; j < n; j++)
      if (i * n + j < l) {
        sm[1][a[j]] += sm[0][a[j]];
        sm[1][a[j]] %= MOD;
        if (i == k - 1) {
          long long mul = (l / n + (j < (l % n)) - k + 1) % MOD;
          tot += (sm[0][a[j]] * mul) % MOD;
          tot %= MOD;
        } else {
          tot += sm[0][a[j]];
          tot %= MOD;
        }
      }
    for (int j = 1; j <= n; j++) {
      sm[1][j] += sm[1][j - 1];
      sm[1][j] %= MOD;
      sm[0][j] = sm[1][j] + 1;
      sm[0][j] %= MOD;
    }
  }
  cout << tot << endl;
  return 0;
}

### Prompt

Please formulate 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 = 1e9 + 7;
int a[1100000];
long long sm[2][1100000];
vector<int> vls;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  long long l;
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    vls.push_back(a[i]);
  }
  sort(vls.begin(), vls.end());
  vector<int>::iterator it;
  it = unique(vls.begin(), vls.end());
  vls.resize(distance(vls.begin(), it));
  for (int i = 0; i < n; i++)
    a[i] = (lower_bound(vls.begin(), vls.end(), a[i]) - vls.begin()) + 1;
  memset(sm, 0, sizeof sm);
  long long tot = 0;
  sm[1][0] = 1;
  sm[0][0] = 1;
  for (int j = 1; j <= n; j++) {
    sm[1][j] += sm[1][j - 1];
    sm[1][j] %= MOD;
    sm[0][j] += sm[1][j];
    sm[0][j] %= MOD;
  }
  for (int i = 0; i < k; i++) {
    for (int j = 0; j <= n; j++) sm[1][j] = 0;
    for (int j = 0; j < n; j++)
      if (i * n + j < l) {
        sm[1][a[j]] += sm[0][a[j]];
        sm[1][a[j]] %= MOD;
        if (i == k - 1) {
          long long mul = (l / n + (j < (l % n)) - k + 1) % MOD;
          tot += (sm[0][a[j]] * mul) % MOD;
          tot %= MOD;
        } else {
          tot += sm[0][a[j]];
          tot %= MOD;
        }
      }
    for (int j = 1; j <= n; j++) {
      sm[1][j] += sm[1][j - 1];
      sm[1][j] %= MOD;
      sm[0][j] = sm[1][j] + 1;
      sm[0][j] %= MOD;
    }
  }
  cout << tot << endl;
  return 0;
}

```