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 MAXN = 1100100;
const long long mod = 1000000007;
struct entry {
  int cnt, last;
  long long nprev;
};
map<int, entry> v;
long long times(long long seq, long long total) {
  return total >= seq ? (total - seq + 1) % mod : 0;
}
int main() {
  cin.sync_with_stdio(false);
  int n, k;
  long long l, rep, rest;
  cin >> n >> l >> k;
  rep = l / n;
  rest = l % n;
  for (int i = 0; i < n; i++) {
    int tmp;
    cin >> tmp;
    v[tmp].cnt++;
    v[tmp].last += i < rest;
  }
  long long total = l % mod, last = 0;
  for (auto &e : v) {
    e.second.nprev = last + e.second.cnt;
    last = e.second.nprev;
  }
  for (int x = 2; x <= k && x <= rep + 1; x++) {
    long long next = 0;
    for (auto &e : v) {
      auto &nr = e.second;
      long long how =
          (nr.last * times(x, rep + 1) + (nr.cnt - nr.last) * times(x, rep)) %
          mod;
      how = (how * nr.nprev) % mod;
      total = (total + how) % mod;
      next = (next + (nr.nprev * nr.cnt) % mod) % mod;
      nr.nprev = next;
    }
  }
  cout << total;
}
