#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, l, k;
  cin >> n >> l >> k;
  long long totRep = l / n;
  long long extra = l % n;
  long long arr[n];
  vector<pair<long long, long long>> myVec;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    myVec.push_back({arr[i], i});
  }
  sort(myVec.begin(), myVec.end());
  vector<long long> numLessEq;
  vector<long long> numLess;
  long long temp = 0, temp2 = 0;
  for (int i = 0; i < n; i++) {
    while (temp < n && myVec[temp].first <= myVec[i].first) temp++;
    while (temp2 < n && myVec[temp2].first < myVec[i].first) temp2++;
    numLessEq.emplace_back(temp);
    numLess.emplace_back(temp2);
  }
  vector<vector<long long>> dp(n, vector<long long>(k + 1));
  vector<vector<long long>> pref(n, vector<long long>(k + 1));
  long long ans = 0;
  for (int j = 1; j <= k; j++) {
    for (int i = 0; i < n; i++) {
      dp[i][j] = 0;
      if (j == 1) {
        dp[i][j] = 1;
      } else {
        dp[i][j] = pref[numLessEq[i] - 1][j - 1];
      }
      pref[i][j] = dp[i][j];
      if (i > 0) pref[i][j] += pref[i - 1][j];
      pref[i][j] %= 1000000007;
      long long toAdd;
      long long temp = dp[i][j];
      if (myVec[i].second < extra) {
        toAdd = max(totRep - j + 2, 0ll) % 1000000007 * temp;
      } else {
        toAdd = max(totRep - j + 1, 0ll) % 1000000007 * temp;
      }
      toAdd %= 1000000007;
      ans += toAdd;
      ans %= 1000000007;
    }
  }
  cout << ans << endl;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, l, k;
  cin >> n >> l >> k;
  long long totRep = l / n;
  long long extra = l % n;
  long long arr[n];
  vector<pair<long long, long long>> myVec;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    myVec.push_back({arr[i], i});
  }
  sort(myVec.begin(), myVec.end());
  vector<long long> numLessEq;
  vector<long long> numLess;
  long long temp = 0, temp2 = 0;
  for (int i = 0; i < n; i++) {
    while (temp < n && myVec[temp].first <= myVec[i].first) temp++;
    while (temp2 < n && myVec[temp2].first < myVec[i].first) temp2++;
    numLessEq.emplace_back(temp);
    numLess.emplace_back(temp2);
  }
  vector<vector<long long>> dp(n, vector<long long>(k + 1));
  vector<vector<long long>> pref(n, vector<long long>(k + 1));
  long long ans = 0;
  for (int j = 1; j <= k; j++) {
    for (int i = 0; i < n; i++) {
      dp[i][j] = 0;
      if (j == 1) {
        dp[i][j] = 1;
      } else {
        dp[i][j] = pref[numLessEq[i] - 1][j - 1];
      }
      pref[i][j] = dp[i][j];
      if (i > 0) pref[i][j] += pref[i - 1][j];
      pref[i][j] %= 1000000007;
      long long toAdd;
      long long temp = dp[i][j];
      if (myVec[i].second < extra) {
        toAdd = max(totRep - j + 2, 0ll) % 1000000007 * temp;
      } else {
        toAdd = max(totRep - j + 1, 0ll) % 1000000007 * temp;
      }
      toAdd %= 1000000007;
      ans += toAdd;
      ans %= 1000000007;
    }
  }
  cout << ans << endl;
}

```