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;
const int mod = 1e9 + 7;
const int lim = 1e6 + 10;
long long m, n;
long long l;
int f[lim];
int g[lim];
long long shengxia;
int need[lim];
vector<long long> sum[lim], sigma[lim];
long long duan;
int main() {
  cin >> m >> l >> n;
  for (int i = 1; i <= m; i++) {
    scanf("%d", &f[i]);
  }
  shengxia = l % m;
  for (int i = 1; i <= shengxia; i++) g[i] = f[i];
  sort(f + 1, f + m + 1);
  sort(g + 1, g + shengxia + 1);
  for (int i = 1; i <= m; i++) {
    need[i - 1] = upper_bound(f + 1, f + m + 1, f[i]) - f - 2;
  }
  long long pre = l / m;
  long long duan = min((long long)n, (long long)pre + (l % m == 0 ? 0 : 1));
  for (int i = 0; i < duan; i++) {
    int len = i + 1;
    for (int j = 0; j < m; j++) {
      if (len == 1)
        sum[len].push_back(1);
      else
        sum[len].push_back(sigma[len - 1][need[j]] % mod);
      if (j == 0)
        sigma[len].push_back(sum[len][j] % mod);
      else
        sigma[len].push_back((sigma[len][j - 1] + sum[len][j]) % mod);
    }
  }
  long long ans = 0;
  for (int i = 1; i <= duan && i <= pre; i++)
    for (int j = 0; j < m; j++) {
      int value = f[i];
      ans = ((long long)ans +
             (long long)(((pre - i + 1) % mod) * sum[i][j] % mod)) %
            mod;
    }
  if (shengxia != 0) {
    ans = (ans + shengxia) % mod;
    for (int i = 1; i <= duan - 1 && i <= pre; i++) {
      for (int j = 1; j <= shengxia; j++) {
        int value = g[j];
        int pos = upper_bound(f + 1, f + m + 1, value) - f - 2;
        ans = ((long long)ans + (long long)(sigma[i][pos])) % mod;
      }
    }
  }
  cout << ans % mod << endl;
}
