#include <bits/stdc++.h>
using namespace std;
int a[1000005];
vector<int> b;
int all[1000005], remain[1000005];
long long dp[1000005];
long long solve(long long a, long long b) { return a * b % 1000000007; }
int main() {
  int n, k;
  long long l;
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    b.push_back(a[i]);
  }
  sort(b.begin(), b.end());
  b.erase(unique(b.begin(), b.end()), b.end());
  for (int i = 0; i < n; i++) {
    a[i] = lower_bound(b.begin(), b.end(), a[i]) - b.begin();
  }
  for (int i = 0; i < n; i++) {
    all[a[i]]++;
  }
  for (int i = 0; i < l % n; i++) {
    remain[a[i]]++;
  }
  int sz = b.size();
  long long lim = l / n;
  dp[0] = 1;
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    for (int j = 0; j < sz - 1; j++) {
      dp[j + 1] = (dp[j] + dp[j + 1]) % 1000000007;
    }
    for (int j = 0; j < sz; j++) {
      if (lim - i > 0) {
        ans = (ans + solve(dp[j],
                           solve(all[j] - remain[j], (lim - i) % 1000000007))) %
              1000000007;
      }
      if (lim - i + 1 > 0) {
        ans =
            (ans + solve(dp[j], solve(remain[j], (lim - i + 1) % 1000000007))) %
            1000000007;
      }
      dp[j] = (dp[j] * all[j]) % 1000000007;
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Generate 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;
int a[1000005];
vector<int> b;
int all[1000005], remain[1000005];
long long dp[1000005];
long long solve(long long a, long long b) { return a * b % 1000000007; }
int main() {
  int n, k;
  long long l;
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    b.push_back(a[i]);
  }
  sort(b.begin(), b.end());
  b.erase(unique(b.begin(), b.end()), b.end());
  for (int i = 0; i < n; i++) {
    a[i] = lower_bound(b.begin(), b.end(), a[i]) - b.begin();
  }
  for (int i = 0; i < n; i++) {
    all[a[i]]++;
  }
  for (int i = 0; i < l % n; i++) {
    remain[a[i]]++;
  }
  int sz = b.size();
  long long lim = l / n;
  dp[0] = 1;
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    for (int j = 0; j < sz - 1; j++) {
      dp[j + 1] = (dp[j] + dp[j + 1]) % 1000000007;
    }
    for (int j = 0; j < sz; j++) {
      if (lim - i > 0) {
        ans = (ans + solve(dp[j],
                           solve(all[j] - remain[j], (lim - i) % 1000000007))) %
              1000000007;
      }
      if (lim - i + 1 > 0) {
        ans =
            (ans + solve(dp[j], solve(remain[j], (lim - i + 1) % 1000000007))) %
            1000000007;
      }
      dp[j] = (dp[j] * all[j]) % 1000000007;
    }
  }
  cout << ans << endl;
  return 0;
}

```