#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
void recalculate(vector<long long>& pv, vector<long long>& dp, vector<int>& v) {
  long long sum = 0;
  for (int i = 0; i < dp.size(); i++) {
    sum = (sum + pv[i]) % mod;
    dp[i] = (sum * 1ll * v[i]) % mod;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, k, l;
  cin >> n >> l >> k;
  vector<int> a(n, 0);
  map<int, int> m;
  m[-1] = 0;
  vector<int> last;
  map<int, int> m2;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    m[a[i]]++;
    if (l % n != 0 && (l - 1) % (long long)n == i) m2 = m;
  }
  vector<int> v;
  for (pair<int, int> i : m) v.push_back(i.second), m2[i.first] += 0;
  for (pair<int, int> i : m2) last.push_back(i.second);
  long long num = min(k, l / n);
  long long ans = 0;
  vector<vector<long long> > dp(num + 1, vector<long long>(v.size(), 0));
  dp[0][0] = 1;
  for (int i = 1; i <= num; i++) recalculate(dp[i - 1], dp[i], v);
  vector<long long> dp_last(v.size(), 0);
  for (int i = 0; i <= min(k - 1, num); i++) {
    dp_last.assign(v.size(), 0);
    recalculate(dp[i], dp_last, last);
    for (int j = 0; j < v.size(); j++) ans = (ans + dp_last[j]) % mod;
  }
  for (int i = 1; i <= num; i++) {
    for (int j = 0; j < v.size(); j++) {
      ans = (ans +
             (((l / n - (long long)(i - 1)) % mod) * 1ll * dp[i][j]) % mod) %
            mod;
    }
  }
  cout << ans << "\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 long long mod = 1e9 + 7;
void recalculate(vector<long long>& pv, vector<long long>& dp, vector<int>& v) {
  long long sum = 0;
  for (int i = 0; i < dp.size(); i++) {
    sum = (sum + pv[i]) % mod;
    dp[i] = (sum * 1ll * v[i]) % mod;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, k, l;
  cin >> n >> l >> k;
  vector<int> a(n, 0);
  map<int, int> m;
  m[-1] = 0;
  vector<int> last;
  map<int, int> m2;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    m[a[i]]++;
    if (l % n != 0 && (l - 1) % (long long)n == i) m2 = m;
  }
  vector<int> v;
  for (pair<int, int> i : m) v.push_back(i.second), m2[i.first] += 0;
  for (pair<int, int> i : m2) last.push_back(i.second);
  long long num = min(k, l / n);
  long long ans = 0;
  vector<vector<long long> > dp(num + 1, vector<long long>(v.size(), 0));
  dp[0][0] = 1;
  for (int i = 1; i <= num; i++) recalculate(dp[i - 1], dp[i], v);
  vector<long long> dp_last(v.size(), 0);
  for (int i = 0; i <= min(k - 1, num); i++) {
    dp_last.assign(v.size(), 0);
    recalculate(dp[i], dp_last, last);
    for (int j = 0; j < v.size(); j++) ans = (ans + dp_last[j]) % mod;
  }
  for (int i = 1; i <= num; i++) {
    for (int j = 0; j < v.size(); j++) {
      ans = (ans +
             (((l / n - (long long)(i - 1)) % mod) * 1ll * dp[i][j]) % mod) %
            mod;
    }
  }
  cout << ans << "\n";
  return 0;
}

```