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>.
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000 * 1000 * 1000 + 7;
int N, K;
long long L;
int add(int a, int b) { return (a + b) % MOD; }
int mul(int a, int b) { return (a * 1LL * b) % MOD; }
int a[(1000000 + 10)];
int b[(1000000 + 10)];
vector<int> dp[(1000000 + 10)];
vector<int> s[(1000000 + 10)];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> L >> K;
  for (int i = 0; i < N; i++) cin >> a[i];
  long long T = L / N;
  T %= MOD;
  int r = L % N;
  for (int i = 0; i <= N; ++i) dp[i].resize(K + 1, 0);
  for (int i = 0; i <= N; ++i) s[i].resize(K + 1, 0);
  copy(a, a + N, b);
  sort(a, a + N);
  for (int i = 1; i <= N; ++i) {
    dp[i][1] = 1;
    s[i][1] = add(dp[i][1], s[i - 1][1]);
  }
  vector<int> t_a(N + 1);
  for (int i = 1; i <= N; i++)
    t_a[i] = lower_bound(a, a + N, a[i - 1] + 1) - a - 1 + 1;
  for (int k = 2; k <= K; k++) {
    for (int i = 1; i <= N; ++i) {
      int t = t_a[i];
      dp[i][k] = s[t][k - 1];
    }
    for (int i = 1; i <= N; ++i) s[i][k] = add(s[i - 1][k], dp[i][k]);
  }
  long long BIG_T = L / N;
  int ans = 0;
  if (L >= N)
    for (int k = 1; k <= K && BIG_T - k + 1 > 0; k++) {
      ans = add(ans, mul(s[N][k], (T - k + 1 + MOD) % MOD));
    }
  for (int i = 0; i < r; ++i)
    t_a[i] = lower_bound(a, a + N, b[i] + 1) - a - 1 + 1;
  if (r != 0) {
    ans = add(ans, r);
    for (int k = 2; k <= K && BIG_T - k + 1 >= 0; k++) {
      for (int i = 0; i < r; i++) {
        int t = t_a[i];
        ans = add(ans, s[t][k - 1]);
      }
    }
  }
  cout << ans << endl;
  return 0;
}
