#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 7;
const long long mod = 1e9 + 7;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int main() {
  ios_base::sync_with_stdio(0);
  int n, k;
  long long l;
  cin >> n >> l >> k;
  vector<int> a(n + 1, 0);
  for (int i = 1; i <= n; i++) cin >> a[i];
  vector<int> ord;
  for (int i = 1; i <= n; i++) ord.push_back(i);
  sort(ord.begin(), ord.end(), [&](int i, int j) { return a[i] < a[j]; });
  vector<vector<long long> > d(k + 1, vector<long long>(n + 1, 0ll));
  long long bl = l / (1ll * n);
  long long bl1 = l % (1ll * n);
  long long res = 0;
  for (int i = 1; i <= n; i++) d[1][i] = 1ll;
  for (int i = 2; i <= k; i++) {
    int h = 0;
    for (int j = 0; j < n; j++) {
      if (j) d[i][ord[j]] = d[i][ord[j - 1]];
      while (h < n && a[ord[h]] <= a[ord[j]]) {
        d[i][ord[j]] += d[i - 1][ord[h]];
        d[i][ord[j]] %= mod;
        h += 1;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= k; j++) {
      if (j <= bl + 1ll && i <= bl1) {
        res += d[j][i] * 1ll * ((bl - j + 2ll) % mod);
      } else if (j <= bl) {
        res += d[j][i] * 1ll * ((bl - j + 1ll) % mod);
      }
      res %= mod;
    }
  }
  cout << res;
}

### Prompt

In Cpp, your task is to solve 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 = 1e5 + 7;
const long long mod = 1e9 + 7;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int main() {
  ios_base::sync_with_stdio(0);
  int n, k;
  long long l;
  cin >> n >> l >> k;
  vector<int> a(n + 1, 0);
  for (int i = 1; i <= n; i++) cin >> a[i];
  vector<int> ord;
  for (int i = 1; i <= n; i++) ord.push_back(i);
  sort(ord.begin(), ord.end(), [&](int i, int j) { return a[i] < a[j]; });
  vector<vector<long long> > d(k + 1, vector<long long>(n + 1, 0ll));
  long long bl = l / (1ll * n);
  long long bl1 = l % (1ll * n);
  long long res = 0;
  for (int i = 1; i <= n; i++) d[1][i] = 1ll;
  for (int i = 2; i <= k; i++) {
    int h = 0;
    for (int j = 0; j < n; j++) {
      if (j) d[i][ord[j]] = d[i][ord[j - 1]];
      while (h < n && a[ord[h]] <= a[ord[j]]) {
        d[i][ord[j]] += d[i - 1][ord[h]];
        d[i][ord[j]] %= mod;
        h += 1;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= k; j++) {
      if (j <= bl + 1ll && i <= bl1) {
        res += d[j][i] * 1ll * ((bl - j + 2ll) % mod);
      } else if (j <= bl) {
        res += d[j][i] * 1ll * ((bl - j + 1ll) % mod);
      }
      res %= mod;
    }
  }
  cout << res;
}

```