#include <bits/stdc++.h>
using namespace std;
long long n, l, k, i, c[1000100], o, d[1000100], ans, yu, f[1000100];
long long tmp, j;
struct qq {
  long long a, b;
  bool operator<(const qq& o) const { return o.a > a; }
} t[1000100];
const long long oo = 1000000007;
int main() {
  scanf("%I64d%I64d%I64d", &n, &l, &k);
  for (i = 0; i < n; ++i) {
    scanf("%I64d", &t[i].a);
    t[i].b = i;
  }
  sort(t, t + n);
  c[0] = 1;
  o = 0;
  for (i = 1; i < n; ++i) {
    if (t[i].a == t[i - 1].a) {
      c[o]++;
    } else {
      c[++o] = 1;
    }
    d[t[i].b] = o;
  }
  ans = l % oo;
  long long len = l / n;
  long long yu = l % n;
  for (i = 0; i <= o; ++i) f[i] = 1;
  for (i = 2; i <= k; ++i) {
    if (len < i) break;
    tmp = 0;
    for (j = 0; j <= o; ++j) {
      tmp = (tmp + (c[j] * f[j]) % oo) % oo;
      f[j] = tmp;
    }
    for (j = 0; j < n; ++j) {
      ans = (ans + (f[d[j]] * ((len - i + 1) % oo)) % oo) % oo;
    }
  }
  for (i = 0; i <= o; ++i) f[i] = 1;
  for (i = 1; i < k; ++i) {
    if (len < i) break;
    tmp = 0;
    for (j = 0; j <= o; ++j) {
      tmp = (tmp + (c[j] * f[j]) % oo) % oo;
      f[j] = tmp;
    }
    for (j = 0; j < yu; ++j) {
      ans = (ans + f[d[j]]) % oo;
    }
  }
  printf("%I64d", 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;
long long n, l, k, i, c[1000100], o, d[1000100], ans, yu, f[1000100];
long long tmp, j;
struct qq {
  long long a, b;
  bool operator<(const qq& o) const { return o.a > a; }
} t[1000100];
const long long oo = 1000000007;
int main() {
  scanf("%I64d%I64d%I64d", &n, &l, &k);
  for (i = 0; i < n; ++i) {
    scanf("%I64d", &t[i].a);
    t[i].b = i;
  }
  sort(t, t + n);
  c[0] = 1;
  o = 0;
  for (i = 1; i < n; ++i) {
    if (t[i].a == t[i - 1].a) {
      c[o]++;
    } else {
      c[++o] = 1;
    }
    d[t[i].b] = o;
  }
  ans = l % oo;
  long long len = l / n;
  long long yu = l % n;
  for (i = 0; i <= o; ++i) f[i] = 1;
  for (i = 2; i <= k; ++i) {
    if (len < i) break;
    tmp = 0;
    for (j = 0; j <= o; ++j) {
      tmp = (tmp + (c[j] * f[j]) % oo) % oo;
      f[j] = tmp;
    }
    for (j = 0; j < n; ++j) {
      ans = (ans + (f[d[j]] * ((len - i + 1) % oo)) % oo) % oo;
    }
  }
  for (i = 0; i <= o; ++i) f[i] = 1;
  for (i = 1; i < k; ++i) {
    if (len < i) break;
    tmp = 0;
    for (j = 0; j <= o; ++j) {
      tmp = (tmp + (c[j] * f[j]) % oo) % oo;
      f[j] = tmp;
    }
    for (j = 0; j < yu; ++j) {
      ans = (ans + f[d[j]]) % oo;
    }
  }
  printf("%I64d", ans);
  return 0;
}

```