#include <bits/stdc++.h>
using namespace std;
int n;
int k;
int o = 1;
long long l;
int a[1000010];
int *p[1000010];
int F[1000010];
int S[1000010];
bool cmp(int *i, int *j) { return *i < *j; }
int main() {
  scanf("%d%I64d%d", &n, &l, &k);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]), p[i] = &a[i];
  sort(p, p + n, cmp);
  p[n] = &a[n];
  for (int i = 0; i < n; *p[i++] = o++)
    while (*p[i + 1] == *p[i]) *p[i++] = o;
  int ans = 0;
  for (int i = 1; i < o; ++i) S[i] = 1;
  for (int i = 0; i < k && i * n < l; ++i) {
    for (int j = 0; j < n; ++j) {
      if (((i)*n + (j)) >= l)
        F[((i)*n + (j))] = 0;
      else
        F[((i)*n + (j))] = S[a[j]];
      if (j < l % n) ans = (ans + S[a[j]]) % 1000000007;
    }
    for (int j = 1; j < o; ++j) S[j] = 0;
    for (int j = 0; j < n; ++j)
      S[a[j]] = (S[a[j]] + F[((i)*n + (j))]) % 1000000007;
    for (int j = 1; j < o; ++j) S[j] = (S[j] + S[j - 1]) % 1000000007;
    ans = (ans + (l / n - i) % 1000000007 * S[o - 1] % 1000000007) % 1000000007;
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
int k;
int o = 1;
long long l;
int a[1000010];
int *p[1000010];
int F[1000010];
int S[1000010];
bool cmp(int *i, int *j) { return *i < *j; }
int main() {
  scanf("%d%I64d%d", &n, &l, &k);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]), p[i] = &a[i];
  sort(p, p + n, cmp);
  p[n] = &a[n];
  for (int i = 0; i < n; *p[i++] = o++)
    while (*p[i + 1] == *p[i]) *p[i++] = o;
  int ans = 0;
  for (int i = 1; i < o; ++i) S[i] = 1;
  for (int i = 0; i < k && i * n < l; ++i) {
    for (int j = 0; j < n; ++j) {
      if (((i)*n + (j)) >= l)
        F[((i)*n + (j))] = 0;
      else
        F[((i)*n + (j))] = S[a[j]];
      if (j < l % n) ans = (ans + S[a[j]]) % 1000000007;
    }
    for (int j = 1; j < o; ++j) S[j] = 0;
    for (int j = 0; j < n; ++j)
      S[a[j]] = (S[a[j]] + F[((i)*n + (j))]) % 1000000007;
    for (int j = 1; j < o; ++j) S[j] = (S[j] + S[j - 1]) % 1000000007;
    ans = (ans + (l / n - i) % 1000000007 * S[o - 1] % 1000000007) % 1000000007;
  }
  printf("%d\n", ans);
  return 0;
}

```