#include <bits/stdc++.h>
using namespace std;
int n, k, t;
long long int l;
int a[1000005], b[1000005];
vector<int> rev[1000005];
vector<long long int> p[1000005], sum2[1000005];
long long int sum[1000005][2];
map<int, int> rnk;
int main() {
  scanf("%d%lld%d", &n, &l, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  sort(b + 1, b + n + 1);
  for (int i = 1; i <= n; i++) {
    if (rnk.count(b[i]) == 0) rnk.insert(pair<int, int>(b[i], i));
  }
  for (int i = 1; i <= n; i++) {
    rev[rnk[a[i]]].push_back(i);
  }
  long long int cse = l / n, rem = l % n;
  if ((long long int)k < cse + 1)
    t = k;
  else
    t = cse + 1;
  for (int i = 1; i <= n; i++) {
    p[i].push_back(0);
    sum2[i].push_back(0);
  }
  bool s = 0;
  for (int j = 1; j <= t; j++) {
    for (int i = 1; i <= n; i++) {
      if (j == 1)
        p[i].push_back(1);
      else
        p[i].push_back(sum[rnk[a[i]]][!s]);
    }
    for (int i = 1; i <= n; i++) {
      sum[i][s] = sum[i - 1][s];
      for (int z = 0; z < rev[i].size(); z++)
        sum[i][s] = (sum[i][s] + p[rev[i][z]][j]) % 1000000007;
    }
    s = !s;
  }
  long long int ans = 0;
  s = 0;
  for (int j = 1; j <= t; j++)
    for (int i = 1; i <= n; i++) {
      if (i == 1)
        sum2[i].push_back(p[i][j] % 1000000007);
      else
        sum2[i].push_back((sum2[i - 1][j] + p[i][j]) % 1000000007);
    }
  for (int i = 1; i <= t; i++) {
    long long int tmp;
    tmp =
        ((sum2[n][i] % 1000000007) * ((cse - i + 1) % 1000000007)) % 1000000007;
    ans = (ans + tmp) % 1000000007;
    if (rem == 0) continue;
    tmp = sum2[rem][i] % 1000000007;
    ans = (ans + tmp) % 1000000007;
  }
  printf("%lld\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;
int n, k, t;
long long int l;
int a[1000005], b[1000005];
vector<int> rev[1000005];
vector<long long int> p[1000005], sum2[1000005];
long long int sum[1000005][2];
map<int, int> rnk;
int main() {
  scanf("%d%lld%d", &n, &l, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  sort(b + 1, b + n + 1);
  for (int i = 1; i <= n; i++) {
    if (rnk.count(b[i]) == 0) rnk.insert(pair<int, int>(b[i], i));
  }
  for (int i = 1; i <= n; i++) {
    rev[rnk[a[i]]].push_back(i);
  }
  long long int cse = l / n, rem = l % n;
  if ((long long int)k < cse + 1)
    t = k;
  else
    t = cse + 1;
  for (int i = 1; i <= n; i++) {
    p[i].push_back(0);
    sum2[i].push_back(0);
  }
  bool s = 0;
  for (int j = 1; j <= t; j++) {
    for (int i = 1; i <= n; i++) {
      if (j == 1)
        p[i].push_back(1);
      else
        p[i].push_back(sum[rnk[a[i]]][!s]);
    }
    for (int i = 1; i <= n; i++) {
      sum[i][s] = sum[i - 1][s];
      for (int z = 0; z < rev[i].size(); z++)
        sum[i][s] = (sum[i][s] + p[rev[i][z]][j]) % 1000000007;
    }
    s = !s;
  }
  long long int ans = 0;
  s = 0;
  for (int j = 1; j <= t; j++)
    for (int i = 1; i <= n; i++) {
      if (i == 1)
        sum2[i].push_back(p[i][j] % 1000000007);
      else
        sum2[i].push_back((sum2[i - 1][j] + p[i][j]) % 1000000007);
    }
  for (int i = 1; i <= t; i++) {
    long long int tmp;
    tmp =
        ((sum2[n][i] % 1000000007) * ((cse - i + 1) % 1000000007)) % 1000000007;
    ans = (ans + tmp) % 1000000007;
    if (rem == 0) continue;
    tmp = sum2[rem][i] % 1000000007;
    ans = (ans + tmp) % 1000000007;
  }
  printf("%lld\n", ans);
  return 0;
}

```