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 a[1010101];
int b[1010101];
int c[1010101], d[1010101];
long long dp[1010101][2];
long long l;
long long n, k;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (i < l % n) c[i] = a[i];
  }
  sort(a, a + n);
  long long len = l % n;
  sort(c, c + len);
  int cntr = len - 1;
  for (int i = n - 1; i >= 0; i--) {
    if (i == n - 1 || a[i] != a[i + 1])
      b[i] = i;
    else
      b[i] = b[i + 1];
    if (cntr >= 0 && c[cntr] == a[i]) {
      d[cntr] = b[i];
      cntr--;
    }
  }
  for (int i = 0; i < n; i++) dp[i][0] = i + 1;
  long long res = l % 1000000007;
  for (long long i = 1; i < k; i++) {
    if ((i + 1) * n <= l) {
      for (int j = 0; j < n; j++) {
        dp[j][i % 2] = dp[b[j]][(i + 1) % 2];
        if (j > 0) dp[j][i % 2] += dp[j - 1][i % 2];
        dp[j][i % 2] %= 1000000007;
      }
      res += (((l / n - i) % 1000000007) * dp[n - 1][i % 2]) % 1000000007;
      res %= 1000000007;
    }
    if (len && i * n < l) {
      long long sum = 0;
      for (int j = 0; j < len; j++) {
        sum += dp[d[j]][(i + 1) % 2];
        sum %= 1000000007;
      }
      res += sum;
      res %= 1000000007;
    }
  }
  cout << res << "\n";
}
