#include <bits/stdc++.h>
using namespace std;
long long const mod = 1e9 + 7;
long long pw(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
long long inv(long long a) { return pw(a, mod - 2); }
long long n, k, s, p[1000010];
long long d[2][1000000];
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> s >> k;
  vector<pair<int, int> > v;
  for (int i = 0; i < n; i++) cin >> p[i], v.push_back(pair<int, int>(p[i], i));
  sort(v.begin(), v.end());
  for (int i = 0; i < n; i++) d[0][i] = 1;
  long long m = s / n, rem = s - m * n;
  long long res = 0;
  for (int i = 1, cur = 0, nxt = 1; i <= k; i++, swap(nxt, cur)) {
    if (m >= i) {
      for (int j = 0; j < n; j++)
        res = (res + ((m + 1 - i) % mod * d[cur][j])) % mod;
    }
    if (m + 1 >= i) {
      for (int j = 0; j < rem; j++) res = (res + (d[cur][j])) % mod;
    } else
      break;
    long long sum = 0;
    for (int l = 0; l < n;) {
      int r = l;
      while (r < n && v[r].first == v[l].first)
        sum += d[cur][v[r++].second], sum %= mod;
      for (int j = l; j < r; j++) d[nxt][v[j].second] = sum;
      l = r;
    }
  }
  cout << res << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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 const mod = 1e9 + 7;
long long pw(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
long long inv(long long a) { return pw(a, mod - 2); }
long long n, k, s, p[1000010];
long long d[2][1000000];
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> s >> k;
  vector<pair<int, int> > v;
  for (int i = 0; i < n; i++) cin >> p[i], v.push_back(pair<int, int>(p[i], i));
  sort(v.begin(), v.end());
  for (int i = 0; i < n; i++) d[0][i] = 1;
  long long m = s / n, rem = s - m * n;
  long long res = 0;
  for (int i = 1, cur = 0, nxt = 1; i <= k; i++, swap(nxt, cur)) {
    if (m >= i) {
      for (int j = 0; j < n; j++)
        res = (res + ((m + 1 - i) % mod * d[cur][j])) % mod;
    }
    if (m + 1 >= i) {
      for (int j = 0; j < rem; j++) res = (res + (d[cur][j])) % mod;
    } else
      break;
    long long sum = 0;
    for (int l = 0; l < n;) {
      int r = l;
      while (r < n && v[r].first == v[l].first)
        sum += d[cur][v[r++].second], sum %= mod;
      for (int j = l; j < r; j++) d[nxt][v[j].second] = sum;
      l = r;
    }
  }
  cout << res << endl;
  return 0;
}

```