#include <bits/stdc++.h>
using namespace std;
unsigned long long n, k, l, total = 0, a[1000000], b[1000000];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> l >> k;
  for (unsigned long long x = 0; x < n; x++) {
    cin >> a[x];
    b[x] = a[x];
  }
  sort(b, b + n);
  vector<vector<unsigned long long> > dp(k, vector<unsigned long long>(n, 0));
  vector<vector<unsigned long long> > ps(k, vector<unsigned long long>(n, 0));
  for (unsigned long long x = 0; x < n; x++) {
    dp[0][x] = 1;
    ps[0][x] = x + 1;
  }
  for (unsigned long long x = 1; x < k; x++) {
    for (unsigned long long y = 0; y < n; y++) {
      dp[x][y] = ps[x - 1][distance(&b[0], upper_bound(b, b + n, b[y])) - 1] %
                 1000000007ll;
      ps[x][y] = dp[x][y] + ((y == 0) ? 0 : ps[x][y - 1]);
      ps[x][y] %= 1000000007ll;
    }
  }
  for (unsigned long long x = 0; x < min(l / n, k); x++) {
    total += ((ps[x][n - 1] % 1000000007ll) *
              ((l / n - x + 1000000007ll) % 1000000007ll)) %
             1000000007ll;
    total %= 1000000007ll;
  }
  for (unsigned long long x = 0; x < (l % n); x++) {
    for (unsigned long long y = 0; y < min(k - 1, l / n); y++) {
      unsigned long long z = distance(&b[0], upper_bound(b, b + n, a[x])) - 1;
      total += ps[y][z];
      total %= 1000000007ll;
    }
  }
  total += l % n;
  total %= 1000000007ll;
  cout << total << '\n';
  return 0;
}

### Prompt

Please create a solution in cpp 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;
unsigned long long n, k, l, total = 0, a[1000000], b[1000000];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> l >> k;
  for (unsigned long long x = 0; x < n; x++) {
    cin >> a[x];
    b[x] = a[x];
  }
  sort(b, b + n);
  vector<vector<unsigned long long> > dp(k, vector<unsigned long long>(n, 0));
  vector<vector<unsigned long long> > ps(k, vector<unsigned long long>(n, 0));
  for (unsigned long long x = 0; x < n; x++) {
    dp[0][x] = 1;
    ps[0][x] = x + 1;
  }
  for (unsigned long long x = 1; x < k; x++) {
    for (unsigned long long y = 0; y < n; y++) {
      dp[x][y] = ps[x - 1][distance(&b[0], upper_bound(b, b + n, b[y])) - 1] %
                 1000000007ll;
      ps[x][y] = dp[x][y] + ((y == 0) ? 0 : ps[x][y - 1]);
      ps[x][y] %= 1000000007ll;
    }
  }
  for (unsigned long long x = 0; x < min(l / n, k); x++) {
    total += ((ps[x][n - 1] % 1000000007ll) *
              ((l / n - x + 1000000007ll) % 1000000007ll)) %
             1000000007ll;
    total %= 1000000007ll;
  }
  for (unsigned long long x = 0; x < (l % n); x++) {
    for (unsigned long long y = 0; y < min(k - 1, l / n); y++) {
      unsigned long long z = distance(&b[0], upper_bound(b, b + n, a[x])) - 1;
      total += ps[y][z];
      total %= 1000000007ll;
    }
  }
  total += l % n;
  total %= 1000000007ll;
  cout << total << '\n';
  return 0;
}

```