#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  char ch = getchar();
  long long x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
struct Node {
  long long x, id;
  inline bool operator<(const Node& rhs) const { return x < rhs.x; }
} a[1000010];
long long f[2][1000010];
long long n, k;
long long l;
inline long long gg(long long len, long long id) {
  long long ret = l / (long long)n + (id < (l % (long long)n));
  ret = max(0ll, (ret - len + 1) % 1000000007ll);
  return ret;
}
int main() {
  n = read();
  l = read();
  k = read();
  for (long long i = (0); i <= (n - 1); i++) a[i].x = read(), a[i].id = i;
  sort(a, a + n);
  long long now = 0;
  long long ans = 0;
  for (long long i = (0); i <= (n - 1); i++)
    f[0][i] = i + 1, (ans += gg(1, a[i].id)) %= 1000000007ll;
  for (long long Case = (2); Case <= (k); Case++) {
    now ^= 1;
    for (long long i = (0); i <= (n - 1); i++) f[now][i] = 0;
    for (long long i = 0; i < n;) {
      long long j = i;
      while (j < n - 1 && a[j + 1].x == a[i].x) j++;
      while (i <= j) {
        f[now][i] = f[now ^ 1][j];
        if (j) f[now][i] = (f[now][i] + f[now][i - 1]) % 1000000007ll;
        (ans += f[now ^ 1][j] * gg(Case, a[i].id) % 1000000007ll) %
            1000000007ll;
        ans %= 1000000007ll;
        i++;
      }
    }
  }
  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;
inline long long read() {
  char ch = getchar();
  long long x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
struct Node {
  long long x, id;
  inline bool operator<(const Node& rhs) const { return x < rhs.x; }
} a[1000010];
long long f[2][1000010];
long long n, k;
long long l;
inline long long gg(long long len, long long id) {
  long long ret = l / (long long)n + (id < (l % (long long)n));
  ret = max(0ll, (ret - len + 1) % 1000000007ll);
  return ret;
}
int main() {
  n = read();
  l = read();
  k = read();
  for (long long i = (0); i <= (n - 1); i++) a[i].x = read(), a[i].id = i;
  sort(a, a + n);
  long long now = 0;
  long long ans = 0;
  for (long long i = (0); i <= (n - 1); i++)
    f[0][i] = i + 1, (ans += gg(1, a[i].id)) %= 1000000007ll;
  for (long long Case = (2); Case <= (k); Case++) {
    now ^= 1;
    for (long long i = (0); i <= (n - 1); i++) f[now][i] = 0;
    for (long long i = 0; i < n;) {
      long long j = i;
      while (j < n - 1 && a[j + 1].x == a[i].x) j++;
      while (i <= j) {
        f[now][i] = f[now ^ 1][j];
        if (j) f[now][i] = (f[now][i] + f[now][i - 1]) % 1000000007ll;
        (ans += f[now ^ 1][j] * gg(Case, a[i].id) % 1000000007ll) %
            1000000007ll;
        ans %= 1000000007ll;
        i++;
      }
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

```