#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e6 + 100, MOD = 1e9 + 7;
int n, k;
long long l, ans, a[MAX], dp[MAX], s[MAX];
pair<int, int> p[MAX];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n * k; i++) p[i] = {a[i % n], i};
  sort(p, p + n * k);
  for (int i = 0; i < n * k; i++) {
    if (p[i].second < n)
      dp[p[i].second] = 1;
    else
      dp[p[i].second] = s[p[i].second / n - 1];
    s[p[i].second / n] = (s[p[i].second / n] + dp[p[i].second]) % MOD;
    if (p[i].second < l)
      ans = (ans +
             ((l - p[i].second - 1) / n + 1) % MOD * dp[p[i].second] % MOD) %
            MOD;
  }
  cout << ans;
}

### Prompt

Please create a solution in CPP 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 MAX = 1e6 + 100, MOD = 1e9 + 7;
int n, k;
long long l, ans, a[MAX], dp[MAX], s[MAX];
pair<int, int> p[MAX];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n * k; i++) p[i] = {a[i % n], i};
  sort(p, p + n * k);
  for (int i = 0; i < n * k; i++) {
    if (p[i].second < n)
      dp[p[i].second] = 1;
    else
      dp[p[i].second] = s[p[i].second / n - 1];
    s[p[i].second / n] = (s[p[i].second / n] + dp[p[i].second]) % MOD;
    if (p[i].second < l)
      ans = (ans +
             ((l - p[i].second - 1) / n + 1) % MOD * dp[p[i].second] % MOD) %
            MOD;
  }
  cout << ans;
}

```