#include <bits/stdc++.h>
using namespace std;
const int NM = 1000010;
const long long MOD = 1000000007;
int N, K, M;
long long L;
long long A[NM];
int Ind[NM];
bool cmpByA(const int& a, const int& b) { return A[a] < A[b]; }
long long DP[NM];
long long ANS;
int main() {
  ios::sync_with_stdio(false);
  cin >> N >> L >> K;
  for (int i = 1; i <= N; i++) {
    cin >> A[i];
    Ind[i] = i;
  }
  sort(Ind + 1, Ind + N + 1, cmpByA);
  for (int i = 1; i <= N; i++) DP[i] = 1;
  ANS = L % MOD;
  for (int k = 2; k <= K; k++) {
    int p = 1;
    long long sum = 0;
    for (int j = 1; j <= N; j++) {
      int i = (k - 1) * N + Ind[j];
      if (1LL * i > L) continue;
      while (p <= N && A[Ind[p]] <= A[Ind[j]]) {
        sum = (sum + DP[(k - 2) * N + Ind[p]]) % MOD;
        p++;
      }
      DP[i] = sum;
      long long cnt = ((L - 1LL * i) / N + 1) % MOD;
      ANS = (1LL * ANS + ((1LL * sum * cnt) % MOD)) % MOD;
    }
  }
  cout << ANS % MOD << '\n';
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

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>.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int NM = 1000010;
const long long MOD = 1000000007;
int N, K, M;
long long L;
long long A[NM];
int Ind[NM];
bool cmpByA(const int& a, const int& b) { return A[a] < A[b]; }
long long DP[NM];
long long ANS;
int main() {
  ios::sync_with_stdio(false);
  cin >> N >> L >> K;
  for (int i = 1; i <= N; i++) {
    cin >> A[i];
    Ind[i] = i;
  }
  sort(Ind + 1, Ind + N + 1, cmpByA);
  for (int i = 1; i <= N; i++) DP[i] = 1;
  ANS = L % MOD;
  for (int k = 2; k <= K; k++) {
    int p = 1;
    long long sum = 0;
    for (int j = 1; j <= N; j++) {
      int i = (k - 1) * N + Ind[j];
      if (1LL * i > L) continue;
      while (p <= N && A[Ind[p]] <= A[Ind[j]]) {
        sum = (sum + DP[(k - 2) * N + Ind[p]]) % MOD;
        p++;
      }
      DP[i] = sum;
      long long cnt = ((L - 1LL * i) / N + 1) % MOD;
      ANS = (1LL * ANS + ((1LL * sum * cnt) % MOD)) % MOD;
    }
  }
  cout << ANS % MOD << '\n';
  return 0;
}

```