#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int N = 1e6 + 100;
long long dp[N];
int main() {
  long long n, l, k;
  while (cin >> n >> l >> k) {
    memset(dp, 0, sizeof(dp));
    map<int, int> f;
    vector<int> vec(n);
    vector<pair<int, int> > q;
    for (int i = 0; i < (n); ++i) scanf("%d", &vec[i]), ++f[vec[i]];
    long long ret = 0;
    for (pair<int, int> top : f) q.push_back(top);
    for (int i = 1; i <= k && i * n <= l; ++i) {
      if (i == 1)
        for (int j = (1); j <= (q.size()); ++j) dp[j] = 1;
      long long tmp = (l / n - i + 1) % MOD;
      for (int j = (1); j <= (q.size()); ++j) {
        dp[j] = dp[j] * q[j - 1].second % MOD;
        (ret += dp[j] * tmp) %= MOD;
        dp[j] = (dp[j] + dp[j - 1]) % MOD;
      }
    }
    memset(dp, 0, sizeof(dp));
    int m = l % n;
    f.clear();
    for (int i = 0; i < (m); ++i) ++f[vec[i]];
    if (m) {
      for (int i = 0; i < k && i * n + m <= l; ++i) {
        if (i == 0) {
          for (pair<int, int> top : f) {
            pair<int, int> tmp = pair<int, int>(top.first, 0);
            int id = lower_bound(q.begin(), q.end(), tmp) - q.begin() + 1;
            dp[id] = top.second;
          }
          ret = (ret + m) % MOD;
          for (int j = (q.size()); j >= (1); --j) (dp[j] += dp[j + 1]) %= MOD;
        } else {
          for (int j = (q.size()); j >= (1); --j) {
            dp[j] = dp[j] * q[j - 1].second % MOD;
            ret = (ret + dp[j]) % MOD;
            dp[j] = (dp[j] + dp[j + 1]) % MOD;
          }
        }
      }
    }
    cout << ret << endl;
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 = 1e9 + 7;
const int N = 1e6 + 100;
long long dp[N];
int main() {
  long long n, l, k;
  while (cin >> n >> l >> k) {
    memset(dp, 0, sizeof(dp));
    map<int, int> f;
    vector<int> vec(n);
    vector<pair<int, int> > q;
    for (int i = 0; i < (n); ++i) scanf("%d", &vec[i]), ++f[vec[i]];
    long long ret = 0;
    for (pair<int, int> top : f) q.push_back(top);
    for (int i = 1; i <= k && i * n <= l; ++i) {
      if (i == 1)
        for (int j = (1); j <= (q.size()); ++j) dp[j] = 1;
      long long tmp = (l / n - i + 1) % MOD;
      for (int j = (1); j <= (q.size()); ++j) {
        dp[j] = dp[j] * q[j - 1].second % MOD;
        (ret += dp[j] * tmp) %= MOD;
        dp[j] = (dp[j] + dp[j - 1]) % MOD;
      }
    }
    memset(dp, 0, sizeof(dp));
    int m = l % n;
    f.clear();
    for (int i = 0; i < (m); ++i) ++f[vec[i]];
    if (m) {
      for (int i = 0; i < k && i * n + m <= l; ++i) {
        if (i == 0) {
          for (pair<int, int> top : f) {
            pair<int, int> tmp = pair<int, int>(top.first, 0);
            int id = lower_bound(q.begin(), q.end(), tmp) - q.begin() + 1;
            dp[id] = top.second;
          }
          ret = (ret + m) % MOD;
          for (int j = (q.size()); j >= (1); --j) (dp[j] += dp[j + 1]) %= MOD;
        } else {
          for (int j = (q.size()); j >= (1); --j) {
            dp[j] = dp[j] * q[j - 1].second % MOD;
            ret = (ret + dp[j]) % MOD;
            dp[j] = (dp[j] + dp[j + 1]) % MOD;
          }
        }
      }
    }
    cout << ret << endl;
  }
  return 0;
}

```