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 MX = 1e6 + 5;
const int MOD = 1000000007;
int n, k, a[MX], aa[MX], pos[MX];
long long l, acc[2][MX], dp[2][MX];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    aa[i] = a[i];
  }
  sort(aa, aa + n);
  int usz = unique(aa, aa + n) - aa;
  for (int i = 0; i < n; i++) pos[i] = lower_bound(aa, aa + usz, a[i]) - aa;
  long long res = 0;
  bool end_processing = false;
  for (int length = 1; length <= k && !end_processing; length++) {
    for (int ending = 0; ending < n; ending++) {
      long long position = ((long long)length - 1) * n + ending + 1;
      if (position > l) {
        end_processing = true;
        break;
      }
      long long rep = (l - position) / n + 1;
      dp[length % 2][ending] = 0;
      if (length == 1) {
        dp[length % 2][ending] = 1;
      } else {
        dp[length % 2][ending] = acc[(length + 1) % 2][pos[ending]];
        dp[length % 2][ending] %= MOD;
      }
      res =
          (res + (((dp[length % 2][ending] % MOD) * (rep % MOD)) % MOD)) % MOD;
    }
    for (int i = 0; i < usz; i++) acc[length % 2][i] = 0;
    for (int i = 0; i < n; i++) {
      acc[length % 2][pos[i]] += dp[length % 2][i];
      acc[length % 2][pos[i]] %= MOD;
    }
    for (int i = 1; i < usz; i++) {
      acc[length % 2][i] = (acc[length % 2][i] + acc[length % 2][i - 1]) % MOD;
    }
  }
  cout << res % MOD << endl;
  return 0;
}
