#include <bits/stdc++.h>
using namespace std;
const int mod = int(1e9) + 7;
int n, k, first[1111111];
long long l;
int a[1111111], p[1111111];
bool cmp(int i, int j) { return a[i] < a[j]; }
int main() {
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    p[i] = i;
  }
  sort(p, p + n, cmp);
  for (int i = 0; i < n; i++) {
    first[i] = 1;
  }
  long long res = 0;
  long long bn = (l - 1) / n;
  for (int j = 1; j <= k; j++) {
    int cur = 0;
    for (int i = 0; i < n; i++) {
      int lst = i + 1;
      while (lst < n && a[p[lst]] == a[p[i]]) {
        ++lst;
      }
      for (int tmp = i; tmp < lst; ++tmp) {
        int ind = p[tmp];
        cur += first[ind];
        if (cur >= mod) cur -= mod;
      }
      while (i < lst) {
        int ind = p[i];
        if (bn + 1 >= j) {
          long long occ = bn + 1 - j;
          if (bn * n + ind < l) {
            ++occ;
          }
          if (occ > 0) {
            occ %= mod;
            res = (res + occ * first[ind]) % mod;
          }
        }
        first[ind] = cur;
        ++i;
      }
      i = lst - 1;
    }
  }
  cout << res << endl;
  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 int mod = int(1e9) + 7;
int n, k, first[1111111];
long long l;
int a[1111111], p[1111111];
bool cmp(int i, int j) { return a[i] < a[j]; }
int main() {
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    p[i] = i;
  }
  sort(p, p + n, cmp);
  for (int i = 0; i < n; i++) {
    first[i] = 1;
  }
  long long res = 0;
  long long bn = (l - 1) / n;
  for (int j = 1; j <= k; j++) {
    int cur = 0;
    for (int i = 0; i < n; i++) {
      int lst = i + 1;
      while (lst < n && a[p[lst]] == a[p[i]]) {
        ++lst;
      }
      for (int tmp = i; tmp < lst; ++tmp) {
        int ind = p[tmp];
        cur += first[ind];
        if (cur >= mod) cur -= mod;
      }
      while (i < lst) {
        int ind = p[i];
        if (bn + 1 >= j) {
          long long occ = bn + 1 - j;
          if (bn * n + ind < l) {
            ++occ;
          }
          if (occ > 0) {
            occ %= mod;
            res = (res + occ * first[ind]) % mod;
          }
        }
        first[ind] = cur;
        ++i;
      }
      i = lst - 1;
    }
  }
  cout << res << endl;
  return 0;
}

```