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, L, K;
long long Array[1000005], Array2[1000005];
const long long MOD = 1000000007;
vector<long long> Matrix[1000005];
char Str[11 * 1000005];
void Read() {
  cin >> N >> L >> K;
  char ch;
  cin.get(ch);
  cin.getline(Str, 11 * 1000005);
  int val = 0;
  int ind = -1;
  for (int i = 0; Str[i] != 0; i++) {
    if (Str[i] >= '0' && Str[i] <= '9')
      val = val * 10 + Str[i] - '0';
    else {
      Array[++ind] = Array2[ind] = val;
      val = 0;
    }
  }
  Array[++ind] = Array2[ind] = val;
  sort(Array, Array + N);
  for (int i = 1; i <= K; i++) Matrix[i].resize(N);
}
int binSearch(int val) {
  int left = 0, right = N - 1, mid, sol = 0;
  while (left <= right) {
    mid = (left + right) / 2;
    if (Array[mid] < val) left = mid + 1;
    if (Array[mid] > val) right = mid - 1;
    if (Array[mid] == val) {
      sol = mid;
      left = mid + 1;
    }
  }
  return sol;
}
int main() {
  Read();
  int n = N, k = K;
  long long number = L / N, remain = L % N, ans = 0;
  long long mod = number % MOD;
  for (int i = 1; i <= k; i++) {
    for (int j = 0; j < n; j++) {
      if (i > number) continue;
      if (i == 1)
        Matrix[i][j] = 1;
      else
        Matrix[i][j] = Matrix[i - 1][binSearch(Array[j])];
      if (j > 0) Matrix[i][j] += Matrix[i][j - 1];
      if (Matrix[i][j] >= MOD) Matrix[i][j] -= MOD;
    }
    ans += Matrix[i][n - 1] * (mod - i + 1);
    ans %= MOD;
    ans += MOD;
    ans %= MOD;
  }
  if (remain > 0) {
    for (int i = 1; i <= k; i++)
      for (int j = 0; j < remain; j++) {
        if (i == 1)
          ++ans;
        else
          ans += Matrix[i - 1][binSearch(Array2[j])];
        if (ans >= MOD) ans -= MOD;
      }
  }
  cout << ans << "\n";
  return 0;
}
