#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.precision(10);
  cout << fixed;
  long long n, l, k;
  cin >> n >> l >> k;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long result = 0;
  long long L = l - (l % n);
  vector<int> b(a);
  vector<int> c(a.begin(), a.begin() + (l % n));
  sort(b.begin(), b.end());
  sort(c.begin(), c.end());
  vector<long long> dp(n, 0), dp_broken(l % n, 0);
  dp[0] = 1;
  long long times = L / n;
  long long times2 = times + (l % n > 0);
  for (int i = 0; i < k; i++) {
    int idx = 0;
    for (int j = 0; j < dp_broken.size(); j++) {
      dp_broken[j] = j ? dp_broken[j - 1] : 0;
      while (idx < n && b[idx] <= c[j]) {
        dp_broken[j] += (dp[idx]);
        if (dp_broken[j] >= MOD) dp_broken[j] -= MOD;
        ;
        idx++;
      }
    }
    if (i < times2) {
      result =
          (result + accumulate(dp_broken.begin(), dp_broken.end(), 0LL)) % MOD;
    }
    idx = 0;
    vector<long long> dp2(n, 0);
    for (int j = 0; j < n; j++) {
      if (j) dp2[j] = dp2[j - 1];
      while (idx < n && b[idx] <= b[j]) {
        dp2[j] += (dp[idx]);
        if (dp2[j] >= MOD) dp2[j] -= MOD;
        ;
        idx++;
      }
    }
    dp = dp2;
    if (i < times) {
      result = (result + (accumulate(dp.begin(), dp.end(), 0LL) % MOD) *
                             ((times - i) % MOD)) %
               MOD;
    }
  }
  cout << result << endl;
}

### 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);
  cin.tie(NULL);
  cout.precision(10);
  cout << fixed;
  long long n, l, k;
  cin >> n >> l >> k;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long result = 0;
  long long L = l - (l % n);
  vector<int> b(a);
  vector<int> c(a.begin(), a.begin() + (l % n));
  sort(b.begin(), b.end());
  sort(c.begin(), c.end());
  vector<long long> dp(n, 0), dp_broken(l % n, 0);
  dp[0] = 1;
  long long times = L / n;
  long long times2 = times + (l % n > 0);
  for (int i = 0; i < k; i++) {
    int idx = 0;
    for (int j = 0; j < dp_broken.size(); j++) {
      dp_broken[j] = j ? dp_broken[j - 1] : 0;
      while (idx < n && b[idx] <= c[j]) {
        dp_broken[j] += (dp[idx]);
        if (dp_broken[j] >= MOD) dp_broken[j] -= MOD;
        ;
        idx++;
      }
    }
    if (i < times2) {
      result =
          (result + accumulate(dp_broken.begin(), dp_broken.end(), 0LL)) % MOD;
    }
    idx = 0;
    vector<long long> dp2(n, 0);
    for (int j = 0; j < n; j++) {
      if (j) dp2[j] = dp2[j - 1];
      while (idx < n && b[idx] <= b[j]) {
        dp2[j] += (dp[idx]);
        if (dp2[j] >= MOD) dp2[j] -= MOD;
        ;
        idx++;
      }
    }
    dp = dp2;
    if (i < times) {
      result = (result + (accumulate(dp.begin(), dp.end(), 0LL) % MOD) *
                             ((times - i) % MOD)) %
               MOD;
    }
  }
  cout << result << endl;
}

```