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 long long MOD = 1000 * 1000 * 1000 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  long long n, l, k;
  long long ans = 0;
  cin >> n >> l >> k;
  vector<long long> A(n);
  for (int i = 0; i < (n); ++i) cin >> A[i];
  vector<vector<long long> > Cnt(k + 1);
  int d;
  unordered_map<long long, int> M;
  {
    int last = -1;
    vector<long long> B(A);
    sort(B.begin(), B.end());
    int cnt = 0;
    for (int i = 0; i < (n); ++i) {
      if (!i || B[i] != B[i - 1]) {
        if (last >= 0) Cnt[1][last] = cnt;
        cnt = 1;
        Cnt[1].push_back(0);
        last++;
        M[B[i]] = last;
      }
      if (i && B[i] == B[i - 1]) cnt++;
    }
    Cnt[1][last] = cnt;
    d = last + 1;
  }
  Cnt[0] = vector<long long>(d, 0);
  Cnt[0][0] = 1;
  for (int(i) = (2); i <= (k); ++(i)) Cnt[i] = vector<long long>(d, 0);
  for (int(i) = (1); i <= (min(k, l / n)); ++(i)) {
    long long pref = 0;
    long long sum = 0;
    for (int(j) = (0); j <= (d - 1); ++(j)) {
      pref = (pref + Cnt[i - 1][j]) % MOD;
      Cnt[i][j] = (pref * Cnt[1][j]) % MOD;
      sum = (sum + Cnt[i][j]) % MOD;
    }
    ans = (ans + (sum % MOD) * ((l / n - i + 1) % MOD)) % MOD;
  }
  vector<long long> CntSuf(d, 0);
  for (int i = 0; i < (l % n); ++i) CntSuf[M[A[i]]]++;
  for (int(i) = (1); i <= (min(k, (l + n - 1) / n)); ++(i)) {
    long long pref = 0;
    for (int(j) = (0); j <= (d - 1); ++(j)) {
      pref = (pref + Cnt[i - 1][j]) % MOD;
      ans = (ans + pref * CntSuf[j]) % MOD;
    }
  }
  cout << ans << endl;
  return 0;
}
