#include <bits/stdc++.h>
using namespace std;
long long v[1000005], a[1000005];
long long n, k, cn;
long long l, ans, aux;
const long long mod = 1e9 + 7;
map<long long, long long> m, poz;
namespace citire {
long long N;
void citire() {
  cin >> n >> l >> k;
  cn = n;
  ans = n * (l / n);
  ans %= mod;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    ++m[a[i]];
    if (m[a[i]] == 1) v[++N] = a[i];
  }
  n = N;
  sort(v + 1, v + n + 1);
}
}  // namespace citire
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  citire::citire();
  long long cl = l, ck = k;
  if (l > cn * k)
    l = cn * k;
  else
    k = l / cn;
  long long dp[k + 5][n + 5], sum[k + 5][n + 5];
  for (int i = 1; i <= k; ++i) sum[i][0] = 0;
  for (int i = 1; i <= n; ++i) {
    dp[1][i] = m[v[i]];
    sum[1][i] = sum[1][i - 1] + dp[1][i];
  }
  for (int p = 2; p <= k; ++p) {
    for (int i = 1; i <= n; ++i) {
      dp[p][i] = sum[p - 1][i] * m[v[i]];
      dp[p][i] %= mod;
      sum[p][i] = sum[p][i - 1] + dp[p][i];
      sum[p][i] %= mod;
      ans += dp[p][i] * ((cl / cn - p + 1) % mod);
      ans %= mod;
    }
  }
  ans %= mod;
  if (cl % cn) {
    for (int i = 1; i <= n; ++i) poz[v[i]] = i;
    n = cl % cn;
    m.clear();
    long long N = 0;
    for (int i = 1; i <= n; ++i) {
      ++m[a[i]];
      if (m[a[i]] == 1) v[++N] = a[i];
    }
    aux = n;
    n = N;
    sort(v + 1, v + n + 1);
    for (int p = 1; p < min(ck, k + 1); ++p) {
      for (int i = 1; i <= n; ++i) {
        aux += (sum[p][poz[v[i]]] * m[v[i]]) % mod;
        aux %= mod;
      }
    }
    aux %= mod;
    ans = (ans + aux) % mod;
  }
  cout << ans;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 v[1000005], a[1000005];
long long n, k, cn;
long long l, ans, aux;
const long long mod = 1e9 + 7;
map<long long, long long> m, poz;
namespace citire {
long long N;
void citire() {
  cin >> n >> l >> k;
  cn = n;
  ans = n * (l / n);
  ans %= mod;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    ++m[a[i]];
    if (m[a[i]] == 1) v[++N] = a[i];
  }
  n = N;
  sort(v + 1, v + n + 1);
}
}  // namespace citire
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  citire::citire();
  long long cl = l, ck = k;
  if (l > cn * k)
    l = cn * k;
  else
    k = l / cn;
  long long dp[k + 5][n + 5], sum[k + 5][n + 5];
  for (int i = 1; i <= k; ++i) sum[i][0] = 0;
  for (int i = 1; i <= n; ++i) {
    dp[1][i] = m[v[i]];
    sum[1][i] = sum[1][i - 1] + dp[1][i];
  }
  for (int p = 2; p <= k; ++p) {
    for (int i = 1; i <= n; ++i) {
      dp[p][i] = sum[p - 1][i] * m[v[i]];
      dp[p][i] %= mod;
      sum[p][i] = sum[p][i - 1] + dp[p][i];
      sum[p][i] %= mod;
      ans += dp[p][i] * ((cl / cn - p + 1) % mod);
      ans %= mod;
    }
  }
  ans %= mod;
  if (cl % cn) {
    for (int i = 1; i <= n; ++i) poz[v[i]] = i;
    n = cl % cn;
    m.clear();
    long long N = 0;
    for (int i = 1; i <= n; ++i) {
      ++m[a[i]];
      if (m[a[i]] == 1) v[++N] = a[i];
    }
    aux = n;
    n = N;
    sort(v + 1, v + n + 1);
    for (int p = 1; p < min(ck, k + 1); ++p) {
      for (int i = 1; i <= n; ++i) {
        aux += (sum[p][poz[v[i]]] * m[v[i]]) % mod;
        aux %= mod;
      }
    }
    aux %= mod;
    ans = (ans + aux) % mod;
  }
  cout << ans;
  return 0;
}

```