#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long a = 0;
  char c = getchar();
  bool f = 0;
  while (!isdigit(c)) {
    f = c == '-';
    c = getchar();
  }
  while (isdigit(c)) {
    a = a * 10 + c - 48;
    c = getchar();
  }
  return f ? -a : a;
}
const long long MOD = 1e9 + 7, _ = 1e6 + 7;
long long dp[2][_], arr[_], id[_], N, L, K, ans;
signed main() {
  N = read();
  L = read();
  K = read();
  ans = L % MOD;
  for (long long i = 0; i < N; ++i) arr[id[i] = i] = read();
  sort(id, id + N, [&](long long p, long long q) { return arr[p] < arr[q]; });
  for (long long i = 0; i < N; ++i) dp[0][i] = 1;
  long long now = 0;
  for (long long i = 2; i <= K; ++i) {
    now ^= 1;
    long long pos = 0, sum = 0;
    for (long long j = 0; j < N; ++j) {
      while (pos < N && arr[id[pos]] == arr[id[j]])
        sum = (sum + dp[now ^ 1][pos++]) % MOD;
      dp[now][j] = sum;
      ans = (ans + max(L / N - i + 1 + (L % N > id[j]), 0ll) % MOD * sum) % MOD;
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
long long read() {
  long long a = 0;
  char c = getchar();
  bool f = 0;
  while (!isdigit(c)) {
    f = c == '-';
    c = getchar();
  }
  while (isdigit(c)) {
    a = a * 10 + c - 48;
    c = getchar();
  }
  return f ? -a : a;
}
const long long MOD = 1e9 + 7, _ = 1e6 + 7;
long long dp[2][_], arr[_], id[_], N, L, K, ans;
signed main() {
  N = read();
  L = read();
  K = read();
  ans = L % MOD;
  for (long long i = 0; i < N; ++i) arr[id[i] = i] = read();
  sort(id, id + N, [&](long long p, long long q) { return arr[p] < arr[q]; });
  for (long long i = 0; i < N; ++i) dp[0][i] = 1;
  long long now = 0;
  for (long long i = 2; i <= K; ++i) {
    now ^= 1;
    long long pos = 0, sum = 0;
    for (long long j = 0; j < N; ++j) {
      while (pos < N && arr[id[pos]] == arr[id[j]])
        sum = (sum + dp[now ^ 1][pos++]) % MOD;
      dp[now][j] = sum;
      ans = (ans + max(L / N - i + 1 + (L % N > id[j]), 0ll) % MOD * sum) % MOD;
    }
  }
  cout << ans;
  return 0;
}

```