#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
pair<int, int> a[1234567];
inline int work(const int &n, const long long &l, const int &kk) {
  int i, j, k;
  int m = min((long long)kk, l / n);
  vector<vector<int> > f(n + 11, vector<int>(m + 11));
  sort(a, a + n);
  for (i = 0; i < (n); i++) f[i][1] = 1;
  for (i = (2); i <= (m); i++) {
    k = 0;
    int sum = 0;
    for (j = 0; j < (n); j++) {
      while (k < n && a[k].first == a[j].first) {
        sum += f[k][i - 1];
        sum %= MOD;
        k++;
      }
      f[j][i] += sum;
    }
  }
  int q = l % n;
  if ((q != 0) && (kk > l / n)) {
    i = m + 1;
    k = 0;
    int sum = 0;
    for (j = 0; j < (n); j++) {
      while (k < n && a[k].first == a[j].first) {
        sum += f[k][i - 1];
        sum %= MOD;
        k++;
      }
      if (a[j].second >= q) continue;
      f[j][i] += sum;
    }
  }
  int ans = 0;
  long long w = l / n;
  for (i = (1); i <= (m + 1); i++) {
    for (j = 0; j < (n); j++) {
      if (f[j][i]) {
        long long multi = w - i + 1;
        if (a[j].second < q) multi++;
        ans += ((multi % MOD) * f[j][i]) % MOD;
        ans %= MOD;
      }
    }
  }
  return ans;
}
int main() {
  int n, k;
  long long l;
  int i;
  cin >> n >> l >> k;
  for (i = 0; i < (n); i++) {
    scanf("%d", &a[i].first);
    a[i].second = i;
  }
  cout << work(n, l, k) << endl;
  return 0;
}

### Prompt

Develop a solution in cpp 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;
const int MOD = 1e9 + 7;
pair<int, int> a[1234567];
inline int work(const int &n, const long long &l, const int &kk) {
  int i, j, k;
  int m = min((long long)kk, l / n);
  vector<vector<int> > f(n + 11, vector<int>(m + 11));
  sort(a, a + n);
  for (i = 0; i < (n); i++) f[i][1] = 1;
  for (i = (2); i <= (m); i++) {
    k = 0;
    int sum = 0;
    for (j = 0; j < (n); j++) {
      while (k < n && a[k].first == a[j].first) {
        sum += f[k][i - 1];
        sum %= MOD;
        k++;
      }
      f[j][i] += sum;
    }
  }
  int q = l % n;
  if ((q != 0) && (kk > l / n)) {
    i = m + 1;
    k = 0;
    int sum = 0;
    for (j = 0; j < (n); j++) {
      while (k < n && a[k].first == a[j].first) {
        sum += f[k][i - 1];
        sum %= MOD;
        k++;
      }
      if (a[j].second >= q) continue;
      f[j][i] += sum;
    }
  }
  int ans = 0;
  long long w = l / n;
  for (i = (1); i <= (m + 1); i++) {
    for (j = 0; j < (n); j++) {
      if (f[j][i]) {
        long long multi = w - i + 1;
        if (a[j].second < q) multi++;
        ans += ((multi % MOD) * f[j][i]) % MOD;
        ans %= MOD;
      }
    }
  }
  return ans;
}
int main() {
  int n, k;
  long long l;
  int i;
  cin >> n >> l >> k;
  for (i = 0; i < (n); i++) {
    scanf("%d", &a[i].first);
    a[i].second = i;
  }
  cout << work(n, l, k) << endl;
  return 0;
}

```