#include <bits/stdc++.h>
using namespace std;
struct node {
  int x, id;
  bool operator<(const node &a) const { return x < a.x; }
} b[1000010];
int n, k;
long long l;
int a[1000010];
long long cnt[1000010], tmp[1000010];
int MOD = 1e9 + 7;
int main(int argc, char const *argv[]) {
  scanf("%d%lld%d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    b[i].x = a[i];
    b[i].id = i;
  }
  if (k == 1) {
    printf("%lld\n", l % MOD);
    return 0;
  }
  sort(b, b + n);
  memset(cnt, 0, sizeof(cnt));
  long long first = 0, second = 0;
  if (n * k < l) {
    first = n;
    for (int i = 0; i < n; i++) cnt[i] = 1;
    for (int i = 1; i < k; i++) {
      tmp[0] = cnt[0];
      for (int j = 1; j < n; j++) tmp[j] = (tmp[j - 1] + cnt[j]) % MOD;
      for (int j = n - 1; j > 0; j--)
        if (b[j].x == b[j - 1].x) tmp[j - 1] = tmp[j];
      for (int j = 0; j < n; j++) first += tmp[j];
      first %= MOD;
      for (int j = 0; j < n; j++) cnt[j] = tmp[j];
    }
    first = first * (((l + n - 1) / n - k) % MOD) % MOD;
  }
  if (l <= n) {
    second = l;
  } else {
    long long tail = (l + n - 1) / n;
    long long head = max(1LL, tail - k + 1);
    for (int i = 0; i < n; i++) cnt[i] = 1;
    second = n;
    for (long long i = head + 1; i <= tail; i++) {
      for (int j = 0; j < n; j++) tmp[j] = (tmp[j - 1] + cnt[j]) % MOD;
      tmp[n - 1]++;
      for (int j = n - 1; j > 0; j--)
        if (b[j].x == b[j - 1].x)
          tmp[j - 1] = tmp[j];
        else
          tmp[j - 1]++;
      if (i != tail)
        for (int j = 0; j < n; j++) second += tmp[j];
      second %= MOD;
      for (int j = 0; j < n; j++) cnt[j] = tmp[j];
    }
    int lim = l - (tail - 1) * n;
    if (head != tail)
      for (int i = 0; i < n; i++)
        if (b[i].id < lim) {
          second += cnt[i];
        }
    second %= MOD;
  }
  printf("%lld\n", (first + second) % MOD);
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
struct node {
  int x, id;
  bool operator<(const node &a) const { return x < a.x; }
} b[1000010];
int n, k;
long long l;
int a[1000010];
long long cnt[1000010], tmp[1000010];
int MOD = 1e9 + 7;
int main(int argc, char const *argv[]) {
  scanf("%d%lld%d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    b[i].x = a[i];
    b[i].id = i;
  }
  if (k == 1) {
    printf("%lld\n", l % MOD);
    return 0;
  }
  sort(b, b + n);
  memset(cnt, 0, sizeof(cnt));
  long long first = 0, second = 0;
  if (n * k < l) {
    first = n;
    for (int i = 0; i < n; i++) cnt[i] = 1;
    for (int i = 1; i < k; i++) {
      tmp[0] = cnt[0];
      for (int j = 1; j < n; j++) tmp[j] = (tmp[j - 1] + cnt[j]) % MOD;
      for (int j = n - 1; j > 0; j--)
        if (b[j].x == b[j - 1].x) tmp[j - 1] = tmp[j];
      for (int j = 0; j < n; j++) first += tmp[j];
      first %= MOD;
      for (int j = 0; j < n; j++) cnt[j] = tmp[j];
    }
    first = first * (((l + n - 1) / n - k) % MOD) % MOD;
  }
  if (l <= n) {
    second = l;
  } else {
    long long tail = (l + n - 1) / n;
    long long head = max(1LL, tail - k + 1);
    for (int i = 0; i < n; i++) cnt[i] = 1;
    second = n;
    for (long long i = head + 1; i <= tail; i++) {
      for (int j = 0; j < n; j++) tmp[j] = (tmp[j - 1] + cnt[j]) % MOD;
      tmp[n - 1]++;
      for (int j = n - 1; j > 0; j--)
        if (b[j].x == b[j - 1].x)
          tmp[j - 1] = tmp[j];
        else
          tmp[j - 1]++;
      if (i != tail)
        for (int j = 0; j < n; j++) second += tmp[j];
      second %= MOD;
      for (int j = 0; j < n; j++) cnt[j] = tmp[j];
    }
    int lim = l - (tail - 1) * n;
    if (head != tail)
      for (int i = 0; i < n; i++)
        if (b[i].id < lim) {
          second += cnt[i];
        }
    second %= MOD;
  }
  printf("%lld\n", (first + second) % MOD);
  return 0;
}

```