#include <bits/stdc++.h>
using namespace std;
set<int> second;
map<int, int> m;
vector<vector<long long int> > dp, suf;
long long int ans = 0;
vector<long long int> temp;
pair<long long int, long long int> ar[1000005];
int main() {
  long long int n, l, k;
  scanf("%lld", &n);
  scanf("%lld", &l);
  scanf("%lld", &k);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &ar[i].first);
    ar[i].second = i;
    second.insert(ar[i].first);
  }
  sort(ar + 1, ar + n + 1);
  int pos = 1;
  for (auto it = second.begin(); it != second.end(); it++) {
    m[*it] = pos;
    pos++;
  }
  for (int i = 0; i <= (int)second.size(); i++) {
    temp.push_back(0);
  }
  for (int i = 0; i <= k; i++) {
    dp.push_back(temp);
    suf.push_back(temp);
  }
  long long int blocks = l / n, blocks2 = l / n;
  long long int t = blocks * n;
  if (l % n != 0) blocks++;
  for (int i = 0; i <= (int)second.size(); i++) {
    suf[0][i] = 1;
  }
  for (int i = 1; i <= k; i++) {
    if (i > blocks) break;
    for (int j = 1; j <= n; j++) {
      long long int temp2 = blocks2;
      if (l - t >= ar[j].second) temp2 = blocks;
      if (temp2 > 0) {
        temp2 = temp2 - i + 1;
        ans = ((ans + ((suf[i - 1][m[ar[j].first]] * ((temp2) % 1000000007)) %
                       1000000007)) %
               1000000007);
        dp[i][m[ar[j].first]] = ((dp[i][m[ar[j].first]] +
                                  ((suf[i - 1][m[ar[j].first]]) % 1000000007)) %
                                 1000000007);
      }
    }
    for (int j = 1; j <= (int)second.size(); j++) {
      suf[i][j] = ((suf[i][j - 1] + dp[i][j]) % 1000000007);
    }
  }
  printf("%lld", ans);
  ;
  printf("\n");
  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;
set<int> second;
map<int, int> m;
vector<vector<long long int> > dp, suf;
long long int ans = 0;
vector<long long int> temp;
pair<long long int, long long int> ar[1000005];
int main() {
  long long int n, l, k;
  scanf("%lld", &n);
  scanf("%lld", &l);
  scanf("%lld", &k);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &ar[i].first);
    ar[i].second = i;
    second.insert(ar[i].first);
  }
  sort(ar + 1, ar + n + 1);
  int pos = 1;
  for (auto it = second.begin(); it != second.end(); it++) {
    m[*it] = pos;
    pos++;
  }
  for (int i = 0; i <= (int)second.size(); i++) {
    temp.push_back(0);
  }
  for (int i = 0; i <= k; i++) {
    dp.push_back(temp);
    suf.push_back(temp);
  }
  long long int blocks = l / n, blocks2 = l / n;
  long long int t = blocks * n;
  if (l % n != 0) blocks++;
  for (int i = 0; i <= (int)second.size(); i++) {
    suf[0][i] = 1;
  }
  for (int i = 1; i <= k; i++) {
    if (i > blocks) break;
    for (int j = 1; j <= n; j++) {
      long long int temp2 = blocks2;
      if (l - t >= ar[j].second) temp2 = blocks;
      if (temp2 > 0) {
        temp2 = temp2 - i + 1;
        ans = ((ans + ((suf[i - 1][m[ar[j].first]] * ((temp2) % 1000000007)) %
                       1000000007)) %
               1000000007);
        dp[i][m[ar[j].first]] = ((dp[i][m[ar[j].first]] +
                                  ((suf[i - 1][m[ar[j].first]]) % 1000000007)) %
                                 1000000007);
      }
    }
    for (int j = 1; j <= (int)second.size(); j++) {
      suf[i][j] = ((suf[i][j - 1] + dp[i][j]) % 1000000007);
    }
  }
  printf("%lld", ans);
  ;
  printf("\n");
  return 0;
}

```