#include <bits/stdc++.h>
using namespace std;
const int ms = 1e6 + 12;
const int mod = 1000000007;
vector<vector<long long>> dp;
long long a[ms], b[ms];
map<long long, int> id;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n, m, k;
  cin >> n >> m >> k;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(a, a + n);
  for (int i = 0; i < n; ++i) {
    id[a[i]] = i;
  }
  dp = vector<vector<long long>>(n + 1, vector<long long>(k + 1, 0ll));
  dp[0][0] = 1;
  for (int j = 1; j <= k; ++j) {
    long long sum = 0, l = 0;
    for (int i = 0; i < n; ++i) {
      while (l < n && a[l] <= a[i]) sum = (sum + dp[l][j - 1]) % mod, l++;
      dp[i][j] = sum;
    }
  }
  long long tk = min(k, m / n);
  long long res = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 1; j <= tk; ++j) {
      res = (res + (dp[i][j] * ((m / n - j + 1) % mod)) % mod) % mod;
    }
  }
  long long top = m % n;
  tk = min(k, m / n + 1);
  for (int i = 0; i < top; ++i) {
    int ti = id[b[i]];
    for (int j = 1; j <= tk; ++j) {
      res = (res + dp[ti][j]) % mod;
    }
  }
  cout << res << "\n";
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
const int ms = 1e6 + 12;
const int mod = 1000000007;
vector<vector<long long>> dp;
long long a[ms], b[ms];
map<long long, int> id;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n, m, k;
  cin >> n >> m >> k;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(a, a + n);
  for (int i = 0; i < n; ++i) {
    id[a[i]] = i;
  }
  dp = vector<vector<long long>>(n + 1, vector<long long>(k + 1, 0ll));
  dp[0][0] = 1;
  for (int j = 1; j <= k; ++j) {
    long long sum = 0, l = 0;
    for (int i = 0; i < n; ++i) {
      while (l < n && a[l] <= a[i]) sum = (sum + dp[l][j - 1]) % mod, l++;
      dp[i][j] = sum;
    }
  }
  long long tk = min(k, m / n);
  long long res = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 1; j <= tk; ++j) {
      res = (res + (dp[i][j] * ((m / n - j + 1) % mod)) % mod) % mod;
    }
  }
  long long top = m % n;
  tk = min(k, m / n + 1);
  for (int i = 0; i < top; ++i) {
    int ti = id[b[i]];
    for (int j = 1; j <= tk; ++j) {
      res = (res + dp[ti][j]) % mod;
    }
  }
  cout << res << "\n";
  return 0;
}

```