#include <bits/stdc++.h>
using std::map;
const int N = 1000000 + 10;
const int mod = 1000000000 + 7;
int n, k;
long long part;
long long l;
int a[N];
int way[N], b[N];
map<int, int> rank;
int R, cnt[N];
int main() {
  std::cin >> n >> l >> k;
  part = (l + n - 1) / n;
  k = std::min(0ll + k, part);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  std::sort(b, b + n);
  for (int i = 0; i < n; i++)
    if (i == 0 || b[i] != b[i - 1]) rank[b[i]] = R++;
  int res = 0;
  for (int i = 1; i <= k; i++) {
    if (i == 1)
      for (int j = 0; j < n; j++) way[j] = 1;
    else
      for (int j = 0; j < n; j++) way[j] = cnt[rank[a[j]]];
    int sum = 0;
    for (int j = 0; j < n; j++) (sum += way[j]) %= mod;
    (res += (1ll * sum * ((part - i + 1 - 1) % mod) % mod)) %= mod;
    int ed = l % n;
    if (ed == 0) ed = n;
    for (int j = 0; j < ed; j++) (res += way[j]) %= mod;
    for (int j = 0; j < n; j++) cnt[j] = 0;
    for (int j = 0; j < n; j++) (cnt[rank[a[j]]] += way[j]) %= mod;
    for (int j = 1; j < R; j++) (cnt[j] += cnt[j - 1]) %= mod;
  }
  printf("%d\n", res);
  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 std::map;
const int N = 1000000 + 10;
const int mod = 1000000000 + 7;
int n, k;
long long part;
long long l;
int a[N];
int way[N], b[N];
map<int, int> rank;
int R, cnt[N];
int main() {
  std::cin >> n >> l >> k;
  part = (l + n - 1) / n;
  k = std::min(0ll + k, part);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  std::sort(b, b + n);
  for (int i = 0; i < n; i++)
    if (i == 0 || b[i] != b[i - 1]) rank[b[i]] = R++;
  int res = 0;
  for (int i = 1; i <= k; i++) {
    if (i == 1)
      for (int j = 0; j < n; j++) way[j] = 1;
    else
      for (int j = 0; j < n; j++) way[j] = cnt[rank[a[j]]];
    int sum = 0;
    for (int j = 0; j < n; j++) (sum += way[j]) %= mod;
    (res += (1ll * sum * ((part - i + 1 - 1) % mod) % mod)) %= mod;
    int ed = l % n;
    if (ed == 0) ed = n;
    for (int j = 0; j < ed; j++) (res += way[j]) %= mod;
    for (int j = 0; j < n; j++) cnt[j] = 0;
    for (int j = 0; j < n; j++) (cnt[rank[a[j]]] += way[j]) %= mod;
    for (int j = 1; j < R; j++) (cnt[j] += cnt[j - 1]) %= mod;
  }
  printf("%d\n", res);
  return 0;
}

```