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() {
  ios_base::sync_with_stdio(false);
  ;
  long long n, l, k;
  cin >> n >> l >> k;
  long long r = (l / n + 1) % 1000000007;
  long long s = (r - 1 + 1000000007) % 1000000007;
  long long x = l % n;
  vector<pair<long long, long long> > v(n);
  for (long long i = 0; i < n; i++) {
    cin >> v[i].first;
    v[i].second = i;
  }
  sort(v.begin(), v.end());
  long long **dp = new long long *[n];
  for (long long i = 0; i < n; i++) dp[i] = new long long[k + 1];
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    dp[i][1] = 1;
    if (v[i].second < x)
      ans = (ans + r) % 1000000007;
    else
      ans = (ans + s) % 1000000007;
  }
  long long sum, id;
  for (long long j = 2; j <= min(k, l / n + 1); j++) {
    sum = 0;
    for (long long i = 0; i < n;) {
      id = i;
      while (id < n && v[id].first == v[i].first) {
        sum = (sum + dp[id][j - 1]) % 1000000007;
        id++;
      }
      id = i;
      while (id < n && v[id].first == v[i].first) {
        dp[id][j] = sum;
        id++;
      }
      i = id;
    }
    for (long long i = 0; i < n; i++) {
      if (v[i].second < x)
        ans = (ans + ((r - j + 1 + 1000000007) % 1000000007 * dp[i][j]) %
                         1000000007) %
              1000000007;
      else
        ans = (ans + ((s - j + 1 + 1000000007) % 1000000007 * dp[i][j]) %
                         1000000007) %
              1000000007;
    }
  }
  cout << ans;
  return 0;
}
