#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, K, a[1001000 + 5], ans = 0, b[1001000 + 5], *f[1001000 + 5],
                          nn[1001000 + 5];
long long L;
int main() {
  cin >> n >> L >> K;
  for (int i = 0; i <= K; ++i) f[i] = new int[n + 5];
  for (int i = 1; i <= n; ++i) a[i] = b[i] = read();
  long long num = L / n;
  int Left = L % n, tot = 1;
  nn[tot] = 1;
  sort(b + 1, b + n + 1);
  for (int i = 2; i <= n; ++nn[tot], ++i)
    if (b[i] != b[i - 1]) b[++tot] = b[i];
  for (int j = 0; j <= tot; ++j) f[0][j] = 1;
  for (int i = 1; i <= K; ++i) {
    f[i][0] = 0;
    for (int j = 1; j <= tot; ++j)
      f[i][j] = 1LL * f[i - 1][j] * nn[j] % 1000000007;
    for (int j = 1; j <= tot; ++j)
      f[i][j] = (f[i][j] + f[i][j - 1]) % 1000000007;
  }
  for (int i = 1; i <= n; ++i) {
    long long N = num + (i <= Left);
    int rk = lower_bound(b + 1, b + tot + 1, a[i]) - b;
    for (int j = 1; j <= min(N, (long long)K); ++j)
      ans =
          (ans + 1LL * f[j - 1][rk] * ((N - j + 1) % 1000000007)) % 1000000007;
  }
  printf("%d\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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, K, a[1001000 + 5], ans = 0, b[1001000 + 5], *f[1001000 + 5],
                          nn[1001000 + 5];
long long L;
int main() {
  cin >> n >> L >> K;
  for (int i = 0; i <= K; ++i) f[i] = new int[n + 5];
  for (int i = 1; i <= n; ++i) a[i] = b[i] = read();
  long long num = L / n;
  int Left = L % n, tot = 1;
  nn[tot] = 1;
  sort(b + 1, b + n + 1);
  for (int i = 2; i <= n; ++nn[tot], ++i)
    if (b[i] != b[i - 1]) b[++tot] = b[i];
  for (int j = 0; j <= tot; ++j) f[0][j] = 1;
  for (int i = 1; i <= K; ++i) {
    f[i][0] = 0;
    for (int j = 1; j <= tot; ++j)
      f[i][j] = 1LL * f[i - 1][j] * nn[j] % 1000000007;
    for (int j = 1; j <= tot; ++j)
      f[i][j] = (f[i][j] + f[i][j - 1]) % 1000000007;
  }
  for (int i = 1; i <= n; ++i) {
    long long N = num + (i <= Left);
    int rk = lower_bound(b + 1, b + tot + 1, a[i]) - b;
    for (int j = 1; j <= min(N, (long long)K); ++j)
      ans =
          (ans + 1LL * f[j - 1][rk] * ((N - j + 1) % 1000000007)) % 1000000007;
  }
  printf("%d\n", ans);
  return 0;
}

```