#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
int main() {
  ios_base ::sync_with_stdio(false);
  int n;
  cin >> n;
  long long l;
  cin >> l;
  int k;
  cin >> k;
  vector<int> v(n, 0);
  long long pieces = l / n;
  int tail = l % n;
  vector<vector<int>> dp(k + 1, vector<int>(n + 1, 0));
  vector<vector<int>> sum(k + 1, vector<int>(n + 1, 0));
  for (int i = 0; i < n; ++i) {
    dp[1][i] = 1;
    sum[1][i] = i + 1;
  }
  vector<pair<int, int>> all;
  for (int i = 0; i < n; ++i) {
    cin >> v[i];
    all.push_back(make_pair(v[i], i));
  }
  sort(all.begin(), all.end());
  vector<int> last(n, 0);
  for (int i = n - 1; i >= 0; --i) {
    if (i == n - 1 || (i < n - 1 && all[i].first != all[i + 1].first))
      last[i] = i;
    else
      last[i] = last[i + 1];
  }
  int ans = 0;
  auto add = [&](int &a, int b) {
    a += b;
    while (a >= MOD) a -= MOD;
    while (a < 0) a += MOD;
  };
  for (int i = 2; i <= k; ++i)
    for (int j = 0; j < n; ++j) {
      add(dp[i][j], sum[i - 1][last[j]]);
      add(sum[i][j], dp[i][j]);
      if (j >= 1) add(sum[i][j], sum[i][j - 1]);
    }
  for (int i = 1; i <= k; ++i)
    for (int j = 0; j < n; ++j) {
      int temp = max(0LL, (pieces - i + 1) % MOD);
      add(ans, 1LL * dp[i][last[j]] * temp % MOD);
    }
  add(ans, tail);
  for (int i = 1; i <= min(1LL * (k - 1), pieces); ++i)
    for (int j = 0; j < n; ++j)
      if (all[j].second < tail) {
        add(ans, sum[i][last[j]]);
      }
  cout << ans << "\n";
}

### 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 int MOD = 1e9 + 7;
int main() {
  ios_base ::sync_with_stdio(false);
  int n;
  cin >> n;
  long long l;
  cin >> l;
  int k;
  cin >> k;
  vector<int> v(n, 0);
  long long pieces = l / n;
  int tail = l % n;
  vector<vector<int>> dp(k + 1, vector<int>(n + 1, 0));
  vector<vector<int>> sum(k + 1, vector<int>(n + 1, 0));
  for (int i = 0; i < n; ++i) {
    dp[1][i] = 1;
    sum[1][i] = i + 1;
  }
  vector<pair<int, int>> all;
  for (int i = 0; i < n; ++i) {
    cin >> v[i];
    all.push_back(make_pair(v[i], i));
  }
  sort(all.begin(), all.end());
  vector<int> last(n, 0);
  for (int i = n - 1; i >= 0; --i) {
    if (i == n - 1 || (i < n - 1 && all[i].first != all[i + 1].first))
      last[i] = i;
    else
      last[i] = last[i + 1];
  }
  int ans = 0;
  auto add = [&](int &a, int b) {
    a += b;
    while (a >= MOD) a -= MOD;
    while (a < 0) a += MOD;
  };
  for (int i = 2; i <= k; ++i)
    for (int j = 0; j < n; ++j) {
      add(dp[i][j], sum[i - 1][last[j]]);
      add(sum[i][j], dp[i][j]);
      if (j >= 1) add(sum[i][j], sum[i][j - 1]);
    }
  for (int i = 1; i <= k; ++i)
    for (int j = 0; j < n; ++j) {
      int temp = max(0LL, (pieces - i + 1) % MOD);
      add(ans, 1LL * dp[i][last[j]] * temp % MOD);
    }
  add(ans, tail);
  for (int i = 1; i <= min(1LL * (k - 1), pieces); ++i)
    for (int j = 0; j < n; ++j)
      if (all[j].second < tail) {
        add(ans, sum[i][last[j]]);
      }
  cout << ans << "\n";
}

```