#include <bits/stdc++.h>
using namespace std;
int n, x;
long long l, m, y, z, k, w, g;
int A[1000005];
int B[1000005], C[1000005];
long long DP[1000005], cnt[1000005], ans;
long long mod = 1000000007ll, tmp;
map<int, int> M;
int main() {
  scanf("%d%I64d%I64d", &n, &l, &k);
  for (int i = (0); i < (n); i++) scanf("%d", &A[i]);
  for (int i = (0); i < (n); i++) B[i] = A[i];
  sort(B, B + n);
  for (int i = (n - 1); i >= (0); i--) {
    if (!M[B[i]]) M[B[i]] = i + 1;
  }
  for (int i = (0); i < (n); i++) C[i] = M[A[i]];
  if (1) {
    for (int i = (0); i < (n); i++) cnt[i] = 1ll;
    for (int i = (0); i < (n); i++) {
      w++;
      tmp = l / (long long)(n);
      if (l % (long long)(n) > (long long)(i)) tmp++;
      tmp %= mod;
      ans = (ans + tmp) % mod;
      if (w == l) {
        printf("%I64d\n", ans);
        return 0;
      }
    }
    for (int i = 2; i <= (int)(k); i++) {
      for (int j = (1); j < (n); j++) cnt[j] = (cnt[j] + cnt[j - 1]) % mod;
      for (int j = (n - 2); j >= (0); j--)
        if (B[j] == B[j + 1]) cnt[j] = cnt[j + 1];
      for (int j = (0); j < (n); j++) DP[j] = cnt[C[j] - 1];
      for (int j = (0); j < (n); j++) {
        w++;
        tmp = l / (long long)(n);
        if (l % (long long)(n) > (long long)(j)) tmp++;
        tmp = tmp - (long long)(i) + 1ll;
        tmp %= mod;
        ans = (ans + (tmp * DP[j]) % mod) % mod;
        if (w == l) {
          printf("%I64d\n", ans);
          return 0;
        }
      }
    }
  }
  printf("%I64d\n", ans);
  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, x;
long long l, m, y, z, k, w, g;
int A[1000005];
int B[1000005], C[1000005];
long long DP[1000005], cnt[1000005], ans;
long long mod = 1000000007ll, tmp;
map<int, int> M;
int main() {
  scanf("%d%I64d%I64d", &n, &l, &k);
  for (int i = (0); i < (n); i++) scanf("%d", &A[i]);
  for (int i = (0); i < (n); i++) B[i] = A[i];
  sort(B, B + n);
  for (int i = (n - 1); i >= (0); i--) {
    if (!M[B[i]]) M[B[i]] = i + 1;
  }
  for (int i = (0); i < (n); i++) C[i] = M[A[i]];
  if (1) {
    for (int i = (0); i < (n); i++) cnt[i] = 1ll;
    for (int i = (0); i < (n); i++) {
      w++;
      tmp = l / (long long)(n);
      if (l % (long long)(n) > (long long)(i)) tmp++;
      tmp %= mod;
      ans = (ans + tmp) % mod;
      if (w == l) {
        printf("%I64d\n", ans);
        return 0;
      }
    }
    for (int i = 2; i <= (int)(k); i++) {
      for (int j = (1); j < (n); j++) cnt[j] = (cnt[j] + cnt[j - 1]) % mod;
      for (int j = (n - 2); j >= (0); j--)
        if (B[j] == B[j + 1]) cnt[j] = cnt[j + 1];
      for (int j = (0); j < (n); j++) DP[j] = cnt[C[j] - 1];
      for (int j = (0); j < (n); j++) {
        w++;
        tmp = l / (long long)(n);
        if (l % (long long)(n) > (long long)(j)) tmp++;
        tmp = tmp - (long long)(i) + 1ll;
        tmp %= mod;
        ans = (ans + (tmp * DP[j]) % mod) % mod;
        if (w == l) {
          printf("%I64d\n", ans);
          return 0;
        }
      }
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

```