#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 10, MOD = 1e9 + 7;
long long n, k, l;
pair<long long, long long> a[MAXN];
inline long long calc(int i, int j) {
  long long id = a[j].second;
  long long res = 0;
  if (id - 1 < (l - 1) % n)
    res = ((l + n - 1) / n - i) % MOD;
  else
    res = (((l - 1) / n) - i) % MOD;
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a, a + n);
  long long dp[k][n];
  for (int i = 0; i < k; i++)
    for (int j = 0; j < n; j++) dp[i][j] = 0;
  for (int i = 0; i < n; i++) dp[0][i] = 1;
  long long ans = l % MOD;
  for (int i = 1; i < k; i++) {
    long long kp = 0;
    long long tmp = 0;
    for (int j = 0; j < n; j++) {
      if (1LL * i * n + a[j].second >= l) continue;
      while (a[kp].first <= a[j].first && kp < n) {
        tmp = (tmp + dp[i - 1][kp]) % MOD;
        kp++;
      }
      dp[i][j] = tmp;
      ans = (ans + dp[i][j] * calc(i, j) % MOD) % MOD;
    }
  }
  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;
const int MAXN = 1e6 + 10, MOD = 1e9 + 7;
long long n, k, l;
pair<long long, long long> a[MAXN];
inline long long calc(int i, int j) {
  long long id = a[j].second;
  long long res = 0;
  if (id - 1 < (l - 1) % n)
    res = ((l + n - 1) / n - i) % MOD;
  else
    res = (((l - 1) / n) - i) % MOD;
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a, a + n);
  long long dp[k][n];
  for (int i = 0; i < k; i++)
    for (int j = 0; j < n; j++) dp[i][j] = 0;
  for (int i = 0; i < n; i++) dp[0][i] = 1;
  long long ans = l % MOD;
  for (int i = 1; i < k; i++) {
    long long kp = 0;
    long long tmp = 0;
    for (int j = 0; j < n; j++) {
      if (1LL * i * n + a[j].second >= l) continue;
      while (a[kp].first <= a[j].first && kp < n) {
        tmp = (tmp + dp[i - 1][kp]) % MOD;
        kp++;
      }
      dp[i][j] = tmp;
      ans = (ans + dp[i][j] * calc(i, j) % MOD) % MOD;
    }
  }
  cout << ans << endl;
  return 0;
}

```