#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, k, l;
  cin >> n >> l >> k;
  vector<long long> v(n), pos(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    pos[i] = i;
  }
  sort(pos.begin(), pos.end(), [&v](int i, int j) { return v[i] < v[j]; });
  vector<vector<long long>> dp(k + 1, vector<long long>(n));
  long long ans = 0;
  fill(dp[1].begin(), dp[1].end(), 1);
  for (int i = 2; i <= k; i++) {
    int lp = 0;
    long long ptot = 0;
    for (int j = 0; j < n; j++) {
      while (lp < n && v[pos[j]] >= v[pos[lp]]) {
        ptot += dp[i - 1][lp];
        lp++;
        ptot %= MOD;
      }
      dp[i][j] = ptot;
    }
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 0; j < n; j++) {
      if ((i - 1) * n + pos[j] < l) {
        long long mult = max(0ll, 1 + (l - 1 - pos[j] - (i - 1) * n) / n);
        mult %= MOD;
        ans += mult * dp[i][j];
        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;
const int MOD = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, k, l;
  cin >> n >> l >> k;
  vector<long long> v(n), pos(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    pos[i] = i;
  }
  sort(pos.begin(), pos.end(), [&v](int i, int j) { return v[i] < v[j]; });
  vector<vector<long long>> dp(k + 1, vector<long long>(n));
  long long ans = 0;
  fill(dp[1].begin(), dp[1].end(), 1);
  for (int i = 2; i <= k; i++) {
    int lp = 0;
    long long ptot = 0;
    for (int j = 0; j < n; j++) {
      while (lp < n && v[pos[j]] >= v[pos[lp]]) {
        ptot += dp[i - 1][lp];
        lp++;
        ptot %= MOD;
      }
      dp[i][j] = ptot;
    }
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 0; j < n; j++) {
      if ((i - 1) * n + pos[j] < l) {
        long long mult = max(0ll, 1 + (l - 1 - pos[j] - (i - 1) * n) / n);
        mult %= MOD;
        ans += mult * dp[i][j];
        ans %= MOD;
      }
    }
  }
  cout << ans;
  return 0;
}

```