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;
struct fff {
  int num;
  int p;
};
fff f[1000001];
struct merges {
  int num;
  int cnt;
  int ncnt;
};
merges m[1000001];
long long sum[1000001];
bool cmp(fff a, fff b) {
  if (a.num == b.num) return a.p < b.p;
  return a.num < b.num;
}
int main() {
  int n, k;
  long long l;
  scanf("%d%I64d%d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &f[i].num);
    f[i].p = i;
  }
  sort(f, f + n, cmp);
  int tot = -1;
  for (int i = 0; i < n; i++) {
    if (i == 0 || f[i].num != f[i - 1].num) {
      tot++;
      m[tot].num = f[i].num;
      m[tot].cnt = 1;
      m[tot].ncnt = 0;
    } else
      m[tot].cnt++;
    if (l % n != 0 && f[i].p >= l % n) m[tot].ncnt++;
  }
  long long ans = l % 1000000007;
  sum[0] = m[0].cnt;
  for (int i = 1; i <= tot; i++) {
    sum[i] = sum[i - 1] + m[i].cnt;
  }
  long long duan = l / n;
  if (l % n) duan++;
  long long maxm = min((long long)k, duan);
  duan = duan % 1000000007;
  for (int i = 2; i <= maxm; i++) {
    for (int j = 0; j <= tot; j++) {
      long long now = (m[j].ncnt * sum[j]) % 1000000007;
      now = (now * (duan - i + 1000000007)) % 1000000007;
      ans = (ans + now) % 1000000007;
      now = ((m[j].cnt - m[j].ncnt) * sum[j]) % 1000000007;
      now = (now * (duan - i + 1 + 1000000007)) % 1000000007;
      ans = (ans + now) % 1000000007;
      sum[j] = (sum[j] * m[j].cnt) % 1000000007;
      if (j) sum[j] = (sum[j] + sum[j - 1]) % 1000000007;
    }
  }
  printf("%I64d\n", ans);
  return 0;
}
