#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
vector<vector<int> > memo;
vector<vector<bool> > used;
vector<int> val, freq;
int mode;
int dp(int last, int cur) {
  if (last == -1) {
    if (mode) return cur == 0;
    return 1;
  }
  if (used[last][cur]) return memo[last][cur];
  used[last][cur] = 1;
  int &dev = memo[last][cur] = dp(last - 1, cur);
  if (cur - 1 >= 0)
    dev = ((long long)dev +
           (long long)freq[last] * (long long)dp(last, cur - 1)) %
          MOD;
  return dev;
}
int main() {
  ios_base ::sync_with_stdio(0);
  int n, K;
  long long L;
  while (cin >> n) {
    cin >> L >> K;
    vector<int> A(n);
    for (int i = 0; i < n; i++) cin >> A[i];
    vector<int> B = A;
    val.clear();
    freq.clear();
    sort(B.begin(), B.end());
    for (int i = 0; i < ((int)(B).size()); i++) {
      int ind = -1;
      for (int j = i; j < ((int)(B).size()); ++j)
        if (B[i] == B[j])
          ind = j;
        else
          break;
      val.push_back(B[i]);
      freq.push_back(ind - i + 1);
      i = ind;
    }
    long long g = L / n;
    memo = vector<vector<int> >(((int)(val).size()) + 5, vector<int>(K + 5));
    used = vector<vector<bool> >(((int)(val).size()) + 5, vector<bool>(K + 5));
    mode = 1;
    int ans = 0;
    for (int len = 1; len <= min(g, (long long)K); ++len)
      ans = ((long long)ans + (long long)((g - len + 1) % MOD) *
                                  (long long)dp(((int)(val).size()) - 1, len)) %
            MOD;
    used = vector<vector<bool> >(((int)(val).size()) + 5, vector<bool>(K + 5));
    mode = 0;
    int lvl = (int)min(g, (long long)K - 1);
    for (int i = 0; i < (int)(L % n); i++) {
      int pos = lower_bound(val.begin(), val.end(), A[i]) - val.begin();
      ans = ans + dp(pos, lvl);
      if (ans >= MOD) ans -= MOD;
    }
    cout << ans << '\n';
  }
}

### Prompt

Please create a solution in Cpp 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 MOD = 1000000007;
vector<vector<int> > memo;
vector<vector<bool> > used;
vector<int> val, freq;
int mode;
int dp(int last, int cur) {
  if (last == -1) {
    if (mode) return cur == 0;
    return 1;
  }
  if (used[last][cur]) return memo[last][cur];
  used[last][cur] = 1;
  int &dev = memo[last][cur] = dp(last - 1, cur);
  if (cur - 1 >= 0)
    dev = ((long long)dev +
           (long long)freq[last] * (long long)dp(last, cur - 1)) %
          MOD;
  return dev;
}
int main() {
  ios_base ::sync_with_stdio(0);
  int n, K;
  long long L;
  while (cin >> n) {
    cin >> L >> K;
    vector<int> A(n);
    for (int i = 0; i < n; i++) cin >> A[i];
    vector<int> B = A;
    val.clear();
    freq.clear();
    sort(B.begin(), B.end());
    for (int i = 0; i < ((int)(B).size()); i++) {
      int ind = -1;
      for (int j = i; j < ((int)(B).size()); ++j)
        if (B[i] == B[j])
          ind = j;
        else
          break;
      val.push_back(B[i]);
      freq.push_back(ind - i + 1);
      i = ind;
    }
    long long g = L / n;
    memo = vector<vector<int> >(((int)(val).size()) + 5, vector<int>(K + 5));
    used = vector<vector<bool> >(((int)(val).size()) + 5, vector<bool>(K + 5));
    mode = 1;
    int ans = 0;
    for (int len = 1; len <= min(g, (long long)K); ++len)
      ans = ((long long)ans + (long long)((g - len + 1) % MOD) *
                                  (long long)dp(((int)(val).size()) - 1, len)) %
            MOD;
    used = vector<vector<bool> >(((int)(val).size()) + 5, vector<bool>(K + 5));
    mode = 0;
    int lvl = (int)min(g, (long long)K - 1);
    for (int i = 0; i < (int)(L % n); i++) {
      int pos = lower_bound(val.begin(), val.end(), A[i]) - val.begin();
      ans = ans + dp(pos, lvl);
      if (ans >= MOD) ans -= MOD;
    }
    cout << ans << '\n';
  }
}

```