#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
const long long INF = 1e9;
const long long MOD = 1000000007;
int main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  long long n, l, k;
  cin >> n >> l >> k;
  long long arr[n];
  vector<pair<long long, long long> > ve;
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
    ve.push_back(make_pair(arr[i], i));
  }
  sort(ve.begin(), ve.end());
  long long dp[k][n];
  memset(dp, 0, sizeof(dp));
  for (long long i = 0; i < n; i++) dp[0][i] = 1;
  for (long long i = 1; i < k; i++) {
    long long num = 0;
    long long p = 0;
    for (long long j = 0; j < n; j++) {
      while (p < n && ve[p].first <= ve[j].first) {
        num += dp[i - 1][p];
        num %= MOD;
        p++;
      }
      dp[i][j] = num;
    }
  }
  long long x = l / n;
  long long rem = l % n;
  long long tot = 0;
  for (long long i = 0; i < k; i++) {
    for (long long j = 0; j < n; j++) {
      if (x >= i && ve[j].second < rem) tot += dp[i][j];
      tot %= MOD;
      tot += ((max(1ll * 0, x - i) % MOD) * dp[i][j]);
      tot %= MOD;
    }
  }
  cout << tot;
}

### Prompt

Please formulate 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 N = 200005;
const long long INF = 1e9;
const long long MOD = 1000000007;
int main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  long long n, l, k;
  cin >> n >> l >> k;
  long long arr[n];
  vector<pair<long long, long long> > ve;
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
    ve.push_back(make_pair(arr[i], i));
  }
  sort(ve.begin(), ve.end());
  long long dp[k][n];
  memset(dp, 0, sizeof(dp));
  for (long long i = 0; i < n; i++) dp[0][i] = 1;
  for (long long i = 1; i < k; i++) {
    long long num = 0;
    long long p = 0;
    for (long long j = 0; j < n; j++) {
      while (p < n && ve[p].first <= ve[j].first) {
        num += dp[i - 1][p];
        num %= MOD;
        p++;
      }
      dp[i][j] = num;
    }
  }
  long long x = l / n;
  long long rem = l % n;
  long long tot = 0;
  for (long long i = 0; i < k; i++) {
    for (long long j = 0; j < n; j++) {
      if (x >= i && ve[j].second < rem) tot += dp[i][j];
      tot %= MOD;
      tot += ((max(1ll * 0, x - i) % MOD) * dp[i][j]);
      tot %= MOD;
    }
  }
  cout << tot;
}

```