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;
int main() {
  long long N, L, K;
  map<int, int> mp, mpe;
  scanf("%lld%lld%lld", &N, &L, &K);
  long long rem = L % N, rc = L / N;
  for (int n = 0; n < int(N); n++) {
    int x;
    scanf("%d", &x);
    mp[x]++;
    if (n < rem) mpe[x]++;
  }
  long long ans = 0;
  vector<long long> res(mp.size(), 1);
  for (int k = 1; k < int(K + 1); k++) {
    int i = 0;
    long long tb = 0;
    for (auto it = mp.begin(); it != mp.end(); ++it, ++i) {
      tb = (tb + res[i] * mpe[it->first]) % 1000000007;
      res[i] = (res[i] * it->second) % 1000000007;
      if (i) res[i] = (res[i] + res[i - 1]) % 1000000007;
    }
    if (k <= rc)
      ans = (ans + res.back() * ((rc - k + 1) % 1000000007)) % 1000000007;
    if (k <= rc + 1) ans = (ans + tb) % 1000000007;
  }
  printf("%lld\n", ans);
}
