#include <bits/stdc++.h>
using namespace std;
pair<int, int> p[1000005];
int main() {
  int n, k;
  long long l;
  scanf("%d%I64d%d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &p[i].first);
    p[i].second = i;
  }
  sort(p, p + n);
  vector<vector<int> > dp(k, vector<int>(n));
  for (int i = 0; i < n; i++) dp[0][i] = 1;
  for (int i = 1; i < k; i++) {
    int sum = 0;
    for (int j = 0, t = 0; j < n; j++) {
      while (t < n && p[t].first <= p[j].first) {
        sum = (sum + dp[i - 1][t]) % 1000000007;
        t++;
      }
      dp[i][j] = sum;
    }
  }
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    for (int j = 0; j < n; j++) {
      long long cnt = l / n + (p[j].second < (l % n));
      if (cnt - i > 0)
        ans = (ans + ((cnt - i) % 1000000007) * (dp[i][j] % 1000000007) %
                         1000000007) %
              1000000007;
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
pair<int, int> p[1000005];
int main() {
  int n, k;
  long long l;
  scanf("%d%I64d%d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &p[i].first);
    p[i].second = i;
  }
  sort(p, p + n);
  vector<vector<int> > dp(k, vector<int>(n));
  for (int i = 0; i < n; i++) dp[0][i] = 1;
  for (int i = 1; i < k; i++) {
    int sum = 0;
    for (int j = 0, t = 0; j < n; j++) {
      while (t < n && p[t].first <= p[j].first) {
        sum = (sum + dp[i - 1][t]) % 1000000007;
        t++;
      }
      dp[i][j] = sum;
    }
  }
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    for (int j = 0; j < n; j++) {
      long long cnt = l / n + (p[j].second < (l % n));
      if (cnt - i > 0)
        ans = (ans + ((cnt - i) % 1000000007) * (dp[i][j] % 1000000007) %
                         1000000007) %
              1000000007;
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

```