#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long full[1000050];
long long diff[1000050];
long long pref[1000050];
long long ar[1000050];
long long rig[1000050];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k, l;
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) cin >> ar[i];
  vector<int> last;
  long long lim = l % n;
  for (int i = 0; i < lim; i++) last.push_back(ar[i]);
  sort(last.begin(), last.end());
  sort(ar, ar + n);
  long long ans = 0;
  long long boxes = l / n;
  for (int i = n - 2; i >= 0; i--) {
    if (ar[i] == ar[i + 1]) rig[i] = rig[i + 1] + 1;
  }
  for (long long i = 0; i < k; i++) {
    if ((i + 1LL) * n > l) break;
    for (long long j = 0; j < n; j++) {
      long long c = i * n;
      c += j;
      if (c - n >= 0)
        full[c] = pref[c - n + rig[j]];
      else
        full[c] = 1;
    }
    pref[(i * n)] = full[i * n];
    for (long long c = (i * n) + 1; c < (i + 1LL) * n; c++) {
      pref[c] += full[c] + pref[c - 1];
      if (pref[c] >= mod) pref[c] %= mod;
    }
    long long c = (i + 1LL) * n;
    if (c <= l && boxes >= i) {
      long long f = boxes - i;
      f %= mod;
      long long cont = (f)*pref[c - 1];
      if (cont >= mod) cont %= mod;
      ans += cont;
      if (ans >= mod) ans -= mod;
    }
    if (c + lim <= l && lim && (i < (k - 1))) {
      int cu = 0;
      int k = i * n;
      for (int j = 0; j < lim; j++) {
        while (cu < n && ar[cu] <= last[j]) cu++;
        cu--;
        ans += pref[k + cu];
        if (ans >= mod) ans -= mod;
      }
    }
  }
  ans += lim;
  if (ans >= mod) ans -= mod;
  cout << ans;
  return 0;
}

### Prompt

Create a solution in cpp for 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 mod = 1e9 + 7;
long long full[1000050];
long long diff[1000050];
long long pref[1000050];
long long ar[1000050];
long long rig[1000050];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k, l;
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) cin >> ar[i];
  vector<int> last;
  long long lim = l % n;
  for (int i = 0; i < lim; i++) last.push_back(ar[i]);
  sort(last.begin(), last.end());
  sort(ar, ar + n);
  long long ans = 0;
  long long boxes = l / n;
  for (int i = n - 2; i >= 0; i--) {
    if (ar[i] == ar[i + 1]) rig[i] = rig[i + 1] + 1;
  }
  for (long long i = 0; i < k; i++) {
    if ((i + 1LL) * n > l) break;
    for (long long j = 0; j < n; j++) {
      long long c = i * n;
      c += j;
      if (c - n >= 0)
        full[c] = pref[c - n + rig[j]];
      else
        full[c] = 1;
    }
    pref[(i * n)] = full[i * n];
    for (long long c = (i * n) + 1; c < (i + 1LL) * n; c++) {
      pref[c] += full[c] + pref[c - 1];
      if (pref[c] >= mod) pref[c] %= mod;
    }
    long long c = (i + 1LL) * n;
    if (c <= l && boxes >= i) {
      long long f = boxes - i;
      f %= mod;
      long long cont = (f)*pref[c - 1];
      if (cont >= mod) cont %= mod;
      ans += cont;
      if (ans >= mod) ans -= mod;
    }
    if (c + lim <= l && lim && (i < (k - 1))) {
      int cu = 0;
      int k = i * n;
      for (int j = 0; j < lim; j++) {
        while (cu < n && ar[cu] <= last[j]) cu++;
        cu--;
        ans += pref[k + cu];
        if (ans >= mod) ans -= mod;
      }
    }
  }
  ans += lim;
  if (ans >= mod) ans -= mod;
  cout << ans;
  return 0;
}

```