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 pow_mod(int b, int p) {
  int r = 1;
  b %= 1000000007;
  while (p) {
    if (p & 1) r = ((int)((((long long)(r)) * (b)) % 1000000007));
    p >>= 1;
    b = ((int)((((long long)(b)) * (b)) % 1000000007));
  }
  return r;
}
int a[1000000], coun[1000000], b[1000000], pos[1000000];
vector<int> V[1000000];
int main() {
  int n, k, i, N, j, res, ch, ch2;
  long long l, m;
  scanf("%d%I64d%d", &n, &l, &k);
  m = l / n;
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  sort(a, a + n);
  N = 0;
  coun[0] = 1;
  for (i = 1; i < n; i++) {
    if (a[i] == a[N]) {
      coun[N]++;
    } else {
      a[++N] = a[i];
      coun[N] = 1;
    }
  }
  N++;
  V[0].push_back(coun[0]);
  for (i = 1; i < k; i++)
    V[0].push_back(
        ((int)((((long long)(coun[0])) * (V[0][i - 1])) % 1000000007)));
  for (j = 1; j < N; j++) {
    V[j].push_back((V[j - 1][0] + coun[j]) % 1000000007);
    for (i = 1; i < k; i++)
      V[j].push_back(
          (V[j - 1][i] +
           ((int)((((long long)(coun[j])) * (V[j][i - 1])) % 1000000007))) %
          1000000007);
  }
  for (i = 0; i < l % n; i++) {
    pos[i] = lower_bound(a, a + N, b[i]) - a;
  }
  res = l % n;
  for (i = 0; i < m && i < k; i++) {
    res = (res + ((int)((((long long)((m - i) % 1000000007)) * (V[N - 1][i])) %
                        1000000007))) %
          1000000007;
    if (i != k - 1)
      for (j = 0; j < l % n; j++) res = (res + V[pos[j]][i]) % 1000000007;
  }
  printf("%d\n", res);
  return 0;
}
