#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 2;
const long long MOD = 1e9 + 7;
int a[N], u[N], tail[N];
map<int, int> to_order;
int count2[N];
long long _d[N], _D[N];
int main() {
  long long ans = 0;
  long long n, k;
  long long l;
  cin >> n >> l >> k;
  long long g = (l - 1) / n + 1;
  long long ntail = l - (g - 1) * n;
  for (int(i) = 1; (i) <= (n); ++(i)) scanf("%d", &a[i]), u[i] = a[i];
  sort(u + 1, u + n + 1);
  int m = unique(u + 1, u + n + 1) - u - 1;
  for (int(i) = 1; (i) <= (m); ++(i)) to_order[u[i]] = i;
  for (int(i) = 1; (i) <= (n); ++(i)) count2[to_order[a[i]]]++;
  for (int(i) = 1; (i) <= (ntail); ++(i)) tail[i] = to_order[a[i]];
  _d[(0) * (m + 1) + (0)] = 1;
  _D[(0) * (m + 1) + (0)] = 1;
  for (int(j) = 1; (j) <= (m); ++(j)) _D[(0) * (m + 1) + (j)] = 1;
  for (int(i) = 1; (i) <= (k); ++(i)) {
    if (i > g) break;
    for (int(j) = 1; (j) <= (m); ++(j)) {
      _d[(i) * (m + 1) + (j)] =
          _D[(i - 1) * (m + 1) + (j)] * (long long)count2[j] % MOD;
      _D[(i) * (m + 1) + (j)] =
          (_D[(i) * (m + 1) + (j - 1)] + _d[(i) * (m + 1) + (j)]) % MOD;
      ans = (ans + (long long)_d[(i) * (m + 1) + (j)] * ((g - i) % MOD)) % MOD;
    }
    for (int(j) = 1; (j) <= (ntail); ++(j))
      ans = (ans + _D[(i - 1) * (m + 1) + (tail[j])]) % MOD;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in cpp for 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 + 2;
const long long MOD = 1e9 + 7;
int a[N], u[N], tail[N];
map<int, int> to_order;
int count2[N];
long long _d[N], _D[N];
int main() {
  long long ans = 0;
  long long n, k;
  long long l;
  cin >> n >> l >> k;
  long long g = (l - 1) / n + 1;
  long long ntail = l - (g - 1) * n;
  for (int(i) = 1; (i) <= (n); ++(i)) scanf("%d", &a[i]), u[i] = a[i];
  sort(u + 1, u + n + 1);
  int m = unique(u + 1, u + n + 1) - u - 1;
  for (int(i) = 1; (i) <= (m); ++(i)) to_order[u[i]] = i;
  for (int(i) = 1; (i) <= (n); ++(i)) count2[to_order[a[i]]]++;
  for (int(i) = 1; (i) <= (ntail); ++(i)) tail[i] = to_order[a[i]];
  _d[(0) * (m + 1) + (0)] = 1;
  _D[(0) * (m + 1) + (0)] = 1;
  for (int(j) = 1; (j) <= (m); ++(j)) _D[(0) * (m + 1) + (j)] = 1;
  for (int(i) = 1; (i) <= (k); ++(i)) {
    if (i > g) break;
    for (int(j) = 1; (j) <= (m); ++(j)) {
      _d[(i) * (m + 1) + (j)] =
          _D[(i - 1) * (m + 1) + (j)] * (long long)count2[j] % MOD;
      _D[(i) * (m + 1) + (j)] =
          (_D[(i) * (m + 1) + (j - 1)] + _d[(i) * (m + 1) + (j)]) % MOD;
      ans = (ans + (long long)_d[(i) * (m + 1) + (j)] * ((g - i) % MOD)) % MOD;
    }
    for (int(j) = 1; (j) <= (ntail); ++(j))
      ans = (ans + _D[(i - 1) * (m + 1) + (tail[j])]) % MOD;
  }
  cout << ans << endl;
  return 0;
}

```