#include <bits/stdc++.h>
using namespace std;
long long n, l, k, ans, c, sum;
long long C[1 << 20], K[1 << 20], A[1 << 23], S[1 << 20];
struct nn {
  long long val;
  bool f;
} N[1 << 20];
bool comp(nn na, nn nb) { return na.val < nb.val; }
int main() {
  long long i, j;
  scanf("%I64d%I64d%I64d", &n, &l, &k);
  for (i = 0; i < n; i++)
    scanf("%I64d", C + i), N[i].val = C[i], N[i].f = i < l % n;
  sort(C, C + n);
  sort(N, N + n, comp);
  for (i = 0; i < n; i++) {
    K[i] = upper_bound(C, C + n, C[i]) - C;
  }
  for (j = 0; j < n; j++) A[(1) * (n + 2) + (j)] = 1;
  for (i = 2; i <= k; i++) {
    S[0] = A[(i - 1) * (n + 2) + (0)];
    for (j = 1; j < n; j++)
      S[j] = (A[(i - 1) * (n + 2) + (j)] + S[j - 1]) % 1000000007;
    for (j = 0; j < n; j++) {
      A[(i) * (n + 2) + (j)] = S[K[j] - 1];
    }
  }
  for (i = 1; i <= k; i++) {
    for (j = 0; j < n; j++) {
      c = 0;
      if ((l / n) >= i) c = (l / n - i + 1) % 1000000007;
      ans = (ans + A[(i) * (n + 2) + (j)] * c) % 1000000007;
    }
  }
  for (j = 0; j < n; j++) {
    if (N[j].f) {
      for (i = 1; i <= k; i++) {
        if (i > l / n + 1) break;
        ans = (ans + A[(i) * (n + 2) + (j)]) % 1000000007;
      }
    }
  }
  printf("%I64d\n", ans);
  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, ans, c, sum;
long long C[1 << 20], K[1 << 20], A[1 << 23], S[1 << 20];
struct nn {
  long long val;
  bool f;
} N[1 << 20];
bool comp(nn na, nn nb) { return na.val < nb.val; }
int main() {
  long long i, j;
  scanf("%I64d%I64d%I64d", &n, &l, &k);
  for (i = 0; i < n; i++)
    scanf("%I64d", C + i), N[i].val = C[i], N[i].f = i < l % n;
  sort(C, C + n);
  sort(N, N + n, comp);
  for (i = 0; i < n; i++) {
    K[i] = upper_bound(C, C + n, C[i]) - C;
  }
  for (j = 0; j < n; j++) A[(1) * (n + 2) + (j)] = 1;
  for (i = 2; i <= k; i++) {
    S[0] = A[(i - 1) * (n + 2) + (0)];
    for (j = 1; j < n; j++)
      S[j] = (A[(i - 1) * (n + 2) + (j)] + S[j - 1]) % 1000000007;
    for (j = 0; j < n; j++) {
      A[(i) * (n + 2) + (j)] = S[K[j] - 1];
    }
  }
  for (i = 1; i <= k; i++) {
    for (j = 0; j < n; j++) {
      c = 0;
      if ((l / n) >= i) c = (l / n - i + 1) % 1000000007;
      ans = (ans + A[(i) * (n + 2) + (j)] * c) % 1000000007;
    }
  }
  for (j = 0; j < n; j++) {
    if (N[j].f) {
      for (i = 1; i <= k; i++) {
        if (i > l / n + 1) break;
        ans = (ans + A[(i) * (n + 2) + (j)]) % 1000000007;
      }
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

```