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 = 1000000;
const long long MOD = 1000000007;
long long n, l, k, m, p, s;
pair<int, bool> a[MAXN];
pair<int, int> b[MAXN];
long long dp[MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> l >> k;
  for (int i = 0; i < n; ++i) {
    cin >> a[i].first;
    a[i].second = (i < l % n);
  }
  sort(a, a + n);
  for (int i = 0; i < n; ++i) {
    if (i > 0 && a[i - 1].first != a[i].first) ++m;
    ++b[m].first;
    b[m].second += a[i].second;
  }
  ++m;
  p = l / n;
  for (int i = 1; i <= k; ++i) {
    for (int j = 0; j < m; ++j) {
      dp[j] = dp[j] * b[j].first % MOD;
      dp[j] = (j > 0) ? (dp[j] + dp[j - 1]) % MOD : dp[j];
      dp[j] = (i == 1) ? 1 : dp[j];
      s = (s + dp[j] * b[j].first % MOD * (max(p - i + 1, 0ll) % MOD) % MOD) %
          MOD;
      s = (s + dp[j] * b[j].second * (p + 1 >= i) % MOD) % MOD;
    }
  }
  cout << s << endl;
  cin >> n;
  return 0;
}
