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, L;
int A[1000010];
int B[1000010];
long long howMany;
long long dp[1000010];
long long pref[1000010];
long long totalAns;
void f1() {
  for (int i = 1; i <= N; i++) {
    pref[i] = pref[i - 1] + 1;
  }
  totalAns += (pref[N] * (howMany % 1000000007)) % 1000000007;
  totalAns %= 1000000007;
  for (int i = 2; i <= min(K, howMany); i++) {
    for (int j = 1; j <= N; j++) {
      int temp = upper_bound(B + 1, B + 1 + N, B[j]) - B;
      dp[j] = pref[temp - 1];
    }
    for (int j = 1; j <= N; j++) {
      pref[j] = pref[j - 1] + dp[j];
      pref[j] %= 1000000007;
    }
    totalAns += (pref[N] * ((howMany - i + 1) % 1000000007)) % 1000000007;
    totalAns %= 1000000007;
  }
}
void f2() {
  for (int i = 1; i <= N; i++) {
    pref[i] = pref[i - 1] + 1;
  }
  totalAns += L % N;
  totalAns %= 1000000007;
  for (int i = 2; i <= min(K, howMany + 1); i++) {
    for (int j = 1; j <= L % N; j++) {
      int temp = upper_bound(B + 1, B + 1 + N, A[j]) - B;
      totalAns += pref[temp - 1];
      totalAns %= 1000000007;
    }
    for (int j = 1; j <= N; j++) {
      int temp = upper_bound(B + 1, B + 1 + N, B[j]) - B;
      dp[j] = pref[temp - 1];
    }
    for (int j = 1; j <= N; j++) {
      pref[j] = pref[j - 1] + dp[j];
      pref[j] %= 1000000007;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> N >> L >> K;
  for (int i = 1; i <= N; i++) {
    cin >> A[i];
    B[i] = A[i];
  }
  sort(B + 1, B + 1 + N);
  howMany = L / N;
  if (howMany) {
    f1();
  }
  if (L % N) {
    f2();
  }
  cout << totalAns << endl;
  return 0;
}
