#include <bits/stdc++.h>
using namespace std;
const int INF = 987654321;
const long long N = 1e6 + 7;
const long long MOD = 1e9 + 7;
vector<int> a;
vector<vector<long long> > f;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(12);
  cout << fixed;
  long long n, l, k;
  cin >> n >> l >> k;
  a = vector<int>(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  vector<int> idx(n);
  for (int i = 0; i < n; ++i) idx[i] = i;
  sort(idx.begin(), idx.end(), [](const int& i, const int& j) {
    return make_pair(a[i], i) < make_pair(a[j], j);
  });
  f = vector<vector<long long> >(k + 1, vector<long long>(n, 0));
  for (int i = 0; i < n; ++i) f[1][i] = 1;
  for (int i = 2; i <= k; ++i) {
    int pointer = 0;
    long long psum = 0;
    for (int j = 0; j < n; ++j) {
      while (pointer < n && a[idx[pointer]] <= a[idx[j]])
        (psum += f[i - 1][idx[pointer++]]) %= MOD;
      f[i][idx[j]] = psum;
    }
  }
  long long res = 0;
  for (int i = 1; i <= min(l / n, k); ++i)
    for (int j = 0; j < n; ++j)
      (res += f[i][j] * ((l / n - i + 1) % MOD)) %= MOD;
  for (int i = 1; i <= min((l + n - 1) / n, k); ++i)
    for (int j = 0; j < l % n; ++j) (res += f[i][j]) %= MOD;
  cout << res << '\n';
  return 0;
}

### 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;
const int INF = 987654321;
const long long N = 1e6 + 7;
const long long MOD = 1e9 + 7;
vector<int> a;
vector<vector<long long> > f;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(12);
  cout << fixed;
  long long n, l, k;
  cin >> n >> l >> k;
  a = vector<int>(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  vector<int> idx(n);
  for (int i = 0; i < n; ++i) idx[i] = i;
  sort(idx.begin(), idx.end(), [](const int& i, const int& j) {
    return make_pair(a[i], i) < make_pair(a[j], j);
  });
  f = vector<vector<long long> >(k + 1, vector<long long>(n, 0));
  for (int i = 0; i < n; ++i) f[1][i] = 1;
  for (int i = 2; i <= k; ++i) {
    int pointer = 0;
    long long psum = 0;
    for (int j = 0; j < n; ++j) {
      while (pointer < n && a[idx[pointer]] <= a[idx[j]])
        (psum += f[i - 1][idx[pointer++]]) %= MOD;
      f[i][idx[j]] = psum;
    }
  }
  long long res = 0;
  for (int i = 1; i <= min(l / n, k); ++i)
    for (int j = 0; j < n; ++j)
      (res += f[i][j] * ((l / n - i + 1) % MOD)) %= MOD;
  for (int i = 1; i <= min((l + n - 1) / n, k); ++i)
    for (int j = 0; j < l % n; ++j) (res += f[i][j]) %= MOD;
  cout << res << '\n';
  return 0;
}

```