#include <bits/stdc++.h>
using namespace ::std;
long long n, l, k;
pair<long long, int> a[1000000 + 5];
vector<long long> ans[1000000 + 5];
long long mk[1000000 + 5];
void init(long long m) {
  mk[0] = 0;
  for (int i = 1; i <= 1000000 && m - i + 1 >= 0; i++)
    mk[i] = (m - i + 1) % 1000000007;
}
void input() {
  scanf("%I64d %I64d %I64d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &a[i].first);
    a[i].second = i;
    ans[i].resize(k + 2);
    ans[i][1] = 1;
  }
  sort(a, a + n);
}
void get_ans() {
  for (int i = 2; i <= k; i++) {
    long long now = 0;
    int p = 0;
    for (int j = 0; j < n; j++) {
      while (p < n && a[j].first >= a[p].first) {
        now = (now + ans[p][i - 1]) % 1000000007;
        p++;
      }
      ans[j][i] = now;
    }
  }
}
long long output() {
  long long m = l / n;
  long long ret = 0;
  int lk = k;
  for (k = 1; k <= lk; k++) {
    long long sum = 0;
    for (int i = 0; i < n; i++) sum = (sum + ans[i][k]) % 1000000007;
    ret = (ret + (sum * mk[k]) % 1000000007) % 1000000007;
    int la = l % n;
    long long now = 0;
    int p = 0;
    if (k - 1 > m) continue;
    for (int i = 0; i < n; i++) {
      while (p < n && a[i].first >= a[p].first) {
        now = (now + ans[p][k - 1]) % 1000000007;
        p++;
      }
      if (k == 1) now = 1;
      if (a[i].second < la) ret = (ret + now) % 1000000007;
    }
  }
  return ret;
}
int main() {
  input();
  init(l / n);
  get_ans();
  printf("%I64d", output());
  return 0;
}

### Prompt

Create a solution in CPP for 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;
long long n, l, k;
pair<long long, int> a[1000000 + 5];
vector<long long> ans[1000000 + 5];
long long mk[1000000 + 5];
void init(long long m) {
  mk[0] = 0;
  for (int i = 1; i <= 1000000 && m - i + 1 >= 0; i++)
    mk[i] = (m - i + 1) % 1000000007;
}
void input() {
  scanf("%I64d %I64d %I64d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &a[i].first);
    a[i].second = i;
    ans[i].resize(k + 2);
    ans[i][1] = 1;
  }
  sort(a, a + n);
}
void get_ans() {
  for (int i = 2; i <= k; i++) {
    long long now = 0;
    int p = 0;
    for (int j = 0; j < n; j++) {
      while (p < n && a[j].first >= a[p].first) {
        now = (now + ans[p][i - 1]) % 1000000007;
        p++;
      }
      ans[j][i] = now;
    }
  }
}
long long output() {
  long long m = l / n;
  long long ret = 0;
  int lk = k;
  for (k = 1; k <= lk; k++) {
    long long sum = 0;
    for (int i = 0; i < n; i++) sum = (sum + ans[i][k]) % 1000000007;
    ret = (ret + (sum * mk[k]) % 1000000007) % 1000000007;
    int la = l % n;
    long long now = 0;
    int p = 0;
    if (k - 1 > m) continue;
    for (int i = 0; i < n; i++) {
      while (p < n && a[i].first >= a[p].first) {
        now = (now + ans[p][k - 1]) % 1000000007;
        p++;
      }
      if (k == 1) now = 1;
      if (a[i].second < la) ret = (ret + now) % 1000000007;
    }
  }
  return ret;
}
int main() {
  input();
  init(l / n);
  get_ans();
  printf("%I64d", output());
  return 0;
}

```