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;
long long n, k, a[3000100], b[3000100];
long long l, ans = 0, f[3000100], g[3000100];
long long safe(long long n) {
  if (n < 0) n = n + ((-n) / 1000000007 + 1) * 1000000007;
  n %= 1000000007;
  return n;
}
int main() {
  scanf("%lld%lld%lld", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    b[i] = a[i];
  }
  sort(b, b + n);
  long long cnt = (l / n + int(l % n != 0));
  long long tmp = l % n;
  long long she = (l / n) % 1000000007;
  if (cnt < k) k = cnt;
  for (long long i = 0; i < k; i++) {
    if (!i) {
      ans += tmp;
      for (int j = 0; j < n; j++) {
        f[j] = 1;
        g[j] = j + 1;
      }
      ans += safe(g[n - 1] * safe(she - i));
      if (ans >= 1000000007) ans %= 1000000007;
    } else {
      for (int j = 0; j < tmp; j++) {
        int ls = 0, rs = n;
        while (ls + 1 < rs) {
          int mid = (ls + rs) / 2;
          if (b[mid] <= a[j])
            ls = mid;
          else
            rs = mid;
        }
        ans += g[ls];
        if (ans >= 1000000007) ans %= 1000000007;
      }
      int k = 0;
      for (int j = 0; j < n; j++) {
        while (k < n - 1 && b[k + 1] <= b[j]) k++;
        f[j] = g[k];
        if (j == 0)
          g[j] = f[j];
        else
          g[j] = safe(g[j - 1] + f[j]);
      }
      ans += safe(g[n - 1] * safe(she - i));
      if (ans >= 1000000007) ans %= 1000000007;
    }
  }
  ans %= 1000000007;
  printf("%lld", ans);
  return 0;
}
