#include <bits/stdc++.h>
using namespace std;
long long l, m;
vector<long long> dp[(1 << 20)];
int n, x, k, cnt, res, a[(1 << 20)], rp[(1 << 20)];
map<int, int> mp, rk;
map<int, int>::iterator it;
void init() {
  for (int i = 0; i <= cnt; i++) dp[i].resize(k + 1);
  dp[0][0] = 1;
  for (int i = 1; i <= cnt; i++)
    for (int j = 0; j <= k; j++)
      if (j == 0)
        dp[i][j] = dp[i - 1][j];
      else
        dp[i][j] = dp[i - 1][j] + dp[i][j - 1] * rp[i] % 1000000007,
        dp[i][j] %= 1000000007;
}
int main() {
  cin >> n >> l >> k, m = l / n;
  for (int i = 0; i < n; i++) scanf("%d", a + i), mp[a[i]]++;
  for (it = mp.begin(); it != mp.end(); it++)
    rk[it->first] = ++cnt, rp[cnt] = it->second;
  init();
  for (int i = 1; i <= min(m, (long long)k); i++)
    res += dp[cnt][i] * ((m - i + 1) % 1000000007) % 1000000007,
        res %= 1000000007;
  for (int i = 0; i < l % n; i++) {
    int s = rk[a[i]];
    for (int j = 0; j <= min(m, (long long)k - 1); j++)
      res += dp[s][j], res %= 1000000007;
  }
  cout << res << endl;
}

### 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;
long long l, m;
vector<long long> dp[(1 << 20)];
int n, x, k, cnt, res, a[(1 << 20)], rp[(1 << 20)];
map<int, int> mp, rk;
map<int, int>::iterator it;
void init() {
  for (int i = 0; i <= cnt; i++) dp[i].resize(k + 1);
  dp[0][0] = 1;
  for (int i = 1; i <= cnt; i++)
    for (int j = 0; j <= k; j++)
      if (j == 0)
        dp[i][j] = dp[i - 1][j];
      else
        dp[i][j] = dp[i - 1][j] + dp[i][j - 1] * rp[i] % 1000000007,
        dp[i][j] %= 1000000007;
}
int main() {
  cin >> n >> l >> k, m = l / n;
  for (int i = 0; i < n; i++) scanf("%d", a + i), mp[a[i]]++;
  for (it = mp.begin(); it != mp.end(); it++)
    rk[it->first] = ++cnt, rp[cnt] = it->second;
  init();
  for (int i = 1; i <= min(m, (long long)k); i++)
    res += dp[cnt][i] * ((m - i + 1) % 1000000007) % 1000000007,
        res %= 1000000007;
  for (int i = 0; i < l % n; i++) {
    int s = rk[a[i]];
    for (int j = 0; j <= min(m, (long long)k - 1); j++)
      res += dp[s][j], res %= 1000000007;
  }
  cout << res << endl;
}

```