#include <bits/stdc++.h>
using namespace std;
const int INF = int(1e9);
const long long INFll = 1ll * INF * INF;
const int MOD = 1000000007;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  int n;
  long long l;
  int k;
  cin >> n >> l >> k;
  vector<pair<int, int> > a(n);
  for (int i = 0; i < n; ++i) cin >> a[i].first, a[i].second = i;
  sort(a.begin(), a.end());
  vector<int> pos(n);
  pos[0] = 0;
  int cur = 0;
  for (int i = 1; i < n; ++i)
    if (a[i].first != a[i - 1].first)
      pos[i] = i, cur = i;
    else
      pos[i] = cur;
  vector<vector<int> > f(k, vector<int>(n));
  for (int i = 0; i < n; ++i) f[0][i] = 1;
  vector<int> add(n);
  for (int i = 0; i < k; ++i) {
    int cadd = 0;
    for (int j = 0; j < n; ++j) {
      cadd = (cadd + add[j]) % MOD;
      f[i][j] = (f[i][j] + cadd) % MOD;
      add[j] = 0;
      add[pos[j]] = (add[pos[j]] + f[i][j]) % MOD;
    }
  }
  int ans = 0;
  for (int i = 0; i < k; ++i) {
    for (int j = 0; j < n; ++j) {
      long long d = l / n;
      int num = l % n - 1;
      d += (num >= a[j].second);
      if (d >= i + 1) ans = (ans + 1ll * (d - i) % MOD * f[i][j]) % MOD;
    }
  }
  cout << ans;
  if (0) cerr << "\nTIME = " << clock() << "\n";
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 INF = int(1e9);
const long long INFll = 1ll * INF * INF;
const int MOD = 1000000007;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  int n;
  long long l;
  int k;
  cin >> n >> l >> k;
  vector<pair<int, int> > a(n);
  for (int i = 0; i < n; ++i) cin >> a[i].first, a[i].second = i;
  sort(a.begin(), a.end());
  vector<int> pos(n);
  pos[0] = 0;
  int cur = 0;
  for (int i = 1; i < n; ++i)
    if (a[i].first != a[i - 1].first)
      pos[i] = i, cur = i;
    else
      pos[i] = cur;
  vector<vector<int> > f(k, vector<int>(n));
  for (int i = 0; i < n; ++i) f[0][i] = 1;
  vector<int> add(n);
  for (int i = 0; i < k; ++i) {
    int cadd = 0;
    for (int j = 0; j < n; ++j) {
      cadd = (cadd + add[j]) % MOD;
      f[i][j] = (f[i][j] + cadd) % MOD;
      add[j] = 0;
      add[pos[j]] = (add[pos[j]] + f[i][j]) % MOD;
    }
  }
  int ans = 0;
  for (int i = 0; i < k; ++i) {
    for (int j = 0; j < n; ++j) {
      long long d = l / n;
      int num = l % n - 1;
      d += (num >= a[j].second);
      if (d >= i + 1) ans = (ans + 1ll * (d - i) % MOD * f[i][j]) % MOD;
    }
  }
  cout << ans;
  if (0) cerr << "\nTIME = " << clock() << "\n";
  return 0;
}

```