#include <bits/stdc++.h>
using namespace std;
int N, k, x;
long long l;
vector<pair<int, int> > a;
int d[1001000], s[1001000], nr[1001000], nr2[1001000];
int poz[1001000];
int main() {
  cin >> N >> l >> k;
  long long L = l / N;
  int M = l % N;
  if (M == 0) {
    --L;
    M = N;
  }
  for (int i = 0; i < N; ++i) {
    scanf("%d", &x);
    a.push_back(make_pair(x, i));
  }
  sort(a.begin(), a.end());
  for (int i = 0; i < N; ++i) {
    poz[a[i].second] = i;
  }
  a.push_back(make_pair(-1, N));
  int curr = 0;
  for (int i = 0; i <= N; ++i) {
    d[i] = 1;
    if (a[i].first != a[curr].first) {
      for (int j = i - 1; j >= curr; --j) {
        nr[j] = i - 1;
      }
      curr = i;
    }
  }
  for (int i = 0; i < M; ++i) {
    nr2[i] = nr[poz[i]];
  }
  long long rez = M;
  int ls = k;
  if (L < ls) ls = L;
  for (int i = 1; i <= ls; ++i) {
    s[0] = d[0];
    for (int j = 1; j < N; ++j) {
      s[j] = (s[j - 1] + d[j]) % 1000000007;
    }
    long long len = (L - i + 1) % 1000000007;
    rez = (rez + 1LL * len * s[N - 1]) % 1000000007;
    if (i < k) {
      for (int j = 0; j < M; ++j) {
        rez = (rez + s[nr2[j]]) % 1000000007;
      }
    }
    for (int j = 0; j < N; ++j) {
      d[j] = s[nr[j]];
    }
  }
  cout << rez << "\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;
int N, k, x;
long long l;
vector<pair<int, int> > a;
int d[1001000], s[1001000], nr[1001000], nr2[1001000];
int poz[1001000];
int main() {
  cin >> N >> l >> k;
  long long L = l / N;
  int M = l % N;
  if (M == 0) {
    --L;
    M = N;
  }
  for (int i = 0; i < N; ++i) {
    scanf("%d", &x);
    a.push_back(make_pair(x, i));
  }
  sort(a.begin(), a.end());
  for (int i = 0; i < N; ++i) {
    poz[a[i].second] = i;
  }
  a.push_back(make_pair(-1, N));
  int curr = 0;
  for (int i = 0; i <= N; ++i) {
    d[i] = 1;
    if (a[i].first != a[curr].first) {
      for (int j = i - 1; j >= curr; --j) {
        nr[j] = i - 1;
      }
      curr = i;
    }
  }
  for (int i = 0; i < M; ++i) {
    nr2[i] = nr[poz[i]];
  }
  long long rez = M;
  int ls = k;
  if (L < ls) ls = L;
  for (int i = 1; i <= ls; ++i) {
    s[0] = d[0];
    for (int j = 1; j < N; ++j) {
      s[j] = (s[j - 1] + d[j]) % 1000000007;
    }
    long long len = (L - i + 1) % 1000000007;
    rez = (rez + 1LL * len * s[N - 1]) % 1000000007;
    if (i < k) {
      for (int j = 0; j < M; ++j) {
        rez = (rez + s[nr2[j]]) % 1000000007;
      }
    }
    for (int j = 0; j < N; ++j) {
      d[j] = s[nr[j]];
    }
  }
  cout << rez << "\n";
  return 0;
}

```