#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 5e5;
long long f[N];
pair<int, int> s[N];
long long l, k, n;
int rk[N];
int main() {
  cin >> n >> l >> k;
  long long t = l / n - (l % n == 0);
  l %= n;
  if (l == 0) l = n;
  for (int i = 0; i < l; ++i) f[i] = 1;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &s[i].first);
    s[i].second = i;
  }
  sort(s, s + n, greater<pair<int, int>>());
  long long ans = l;
  for (int i = 1; i <= min(k - 1, t); ++i) {
    long long sum = 0;
    for (int j = 0; j < n; ++j) {
      int k = j;
      for (; k < n && s[j].first == s[k].first; ++k)
        sum = (sum + f[(i - 1) * n + s[k].second]) % 1000000007;
      k = j;
      for (; k < n && s[j].first == s[k].first; ++k) {
        f[i * n + s[k].second] = sum;
        ans = (ans + sum) % 1000000007;
      }
      j = k - 1;
    }
  }
  if (t > 0) {
    for (int i = 0; i < n; ++i) f[i] = 1;
    ans = (ans + n * t) % 1000000007;
    for (int i = 1; i < min(t, k); ++i) {
      long long sum = 0;
      for (int j = 0; j < n; ++j) {
        int k = j;
        for (; k < n && s[j].first == s[k].first; ++k)
          sum = (sum + f[(i - 1) * n + s[k].second]) % 1000000007;
        k = j;
        for (; k < n && s[j].first == s[k].first; ++k) {
          f[i * n + s[k].second] = sum;
          ans = (ans + sum * ((t - i) % 1000000007)) % 1000000007;
        }
        j = k - 1;
      }
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your challenge is to write 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 N = 2e6 + 5e5;
long long f[N];
pair<int, int> s[N];
long long l, k, n;
int rk[N];
int main() {
  cin >> n >> l >> k;
  long long t = l / n - (l % n == 0);
  l %= n;
  if (l == 0) l = n;
  for (int i = 0; i < l; ++i) f[i] = 1;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &s[i].first);
    s[i].second = i;
  }
  sort(s, s + n, greater<pair<int, int>>());
  long long ans = l;
  for (int i = 1; i <= min(k - 1, t); ++i) {
    long long sum = 0;
    for (int j = 0; j < n; ++j) {
      int k = j;
      for (; k < n && s[j].first == s[k].first; ++k)
        sum = (sum + f[(i - 1) * n + s[k].second]) % 1000000007;
      k = j;
      for (; k < n && s[j].first == s[k].first; ++k) {
        f[i * n + s[k].second] = sum;
        ans = (ans + sum) % 1000000007;
      }
      j = k - 1;
    }
  }
  if (t > 0) {
    for (int i = 0; i < n; ++i) f[i] = 1;
    ans = (ans + n * t) % 1000000007;
    for (int i = 1; i < min(t, k); ++i) {
      long long sum = 0;
      for (int j = 0; j < n; ++j) {
        int k = j;
        for (; k < n && s[j].first == s[k].first; ++k)
          sum = (sum + f[(i - 1) * n + s[k].second]) % 1000000007;
        k = j;
        for (; k < n && s[j].first == s[k].first; ++k) {
          f[i * n + s[k].second] = sum;
          ans = (ans + sum * ((t - i) % 1000000007)) % 1000000007;
        }
        j = k - 1;
      }
    }
  }
  cout << ans << endl;
  return 0;
}

```