#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1000000007;
const long double PI = 2 * acos(0.0);
template <typename T = long long int>
vector<T> ga(long long int n, bool oneIndexed = false) {
  vector<T> a = vector<T>(n + oneIndexed);
  for (long long int i = 0; i < n; i++) {
    T p;
    cin >> p;
    a[i + oneIndexed] = p;
  }
  return move(a);
}
template <typename T, typename A>
void pa(vector<T, A> const &a, long long int begin = 0,
        long long int end = -1) {
  if (end == -1) end = (long long int)a.size() - 1;
  for (long long int i = begin; i <= end; i++) {
    cout << a[i] << " ";
  }
  cout << "\n";
}
void solve() {
  long long int n, l, k;
  cin >> n >> l >> k;
  vector<vector<long long int> > dp(n + 10, vector<long long int>(k + 10, 0));
  vector<long long int> a = ga(n);
  for (long long int i = 0; i <= n; i++) {
    dp[i][1] = 1;
  }
  vector<pair<long long int, long long int> > arr;
  for (long long int i = 0; i < n; i++) {
    arr.push_back({a[i], i});
  }
  sort(arr.begin(), arr.end());
  for (long long int j = 2; j <= k; j++) {
    long long int sum = 0, ptr = 0;
    for (long long int i = 0; i < n; i++) {
      while (ptr < n && a[arr[ptr].second] <= a[arr[i].second]) {
        sum = (sum + dp[arr[ptr].second][j - 1]) % MOD;
        ptr++;
      }
      dp[arr[i].second][j] = sum % MOD;
    }
  }
  long long int c = ((l + n - 1) / n), x = (l - 1) % n;
  long long int ans = 0;
  for (long long int i = 0; i <= x; i++) {
    for (long long int j = 1; j <= min(c, k); j++) {
      ans = (ans + (dp[i][j] * ((c - j + 1 + MOD) % MOD)) % MOD) % MOD;
    }
  }
  for (long long int i = x + 1; i < n; i++) {
    for (long long int j = 1; j <= min(c - 1, k); j++) {
      ans = (ans + (dp[i][j] * ((c - j + MOD) % MOD)) % MOD) % MOD;
    }
  }
  cout << ans << '\n';
}
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  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 long long int MOD = 1000000007;
const long double PI = 2 * acos(0.0);
template <typename T = long long int>
vector<T> ga(long long int n, bool oneIndexed = false) {
  vector<T> a = vector<T>(n + oneIndexed);
  for (long long int i = 0; i < n; i++) {
    T p;
    cin >> p;
    a[i + oneIndexed] = p;
  }
  return move(a);
}
template <typename T, typename A>
void pa(vector<T, A> const &a, long long int begin = 0,
        long long int end = -1) {
  if (end == -1) end = (long long int)a.size() - 1;
  for (long long int i = begin; i <= end; i++) {
    cout << a[i] << " ";
  }
  cout << "\n";
}
void solve() {
  long long int n, l, k;
  cin >> n >> l >> k;
  vector<vector<long long int> > dp(n + 10, vector<long long int>(k + 10, 0));
  vector<long long int> a = ga(n);
  for (long long int i = 0; i <= n; i++) {
    dp[i][1] = 1;
  }
  vector<pair<long long int, long long int> > arr;
  for (long long int i = 0; i < n; i++) {
    arr.push_back({a[i], i});
  }
  sort(arr.begin(), arr.end());
  for (long long int j = 2; j <= k; j++) {
    long long int sum = 0, ptr = 0;
    for (long long int i = 0; i < n; i++) {
      while (ptr < n && a[arr[ptr].second] <= a[arr[i].second]) {
        sum = (sum + dp[arr[ptr].second][j - 1]) % MOD;
        ptr++;
      }
      dp[arr[i].second][j] = sum % MOD;
    }
  }
  long long int c = ((l + n - 1) / n), x = (l - 1) % n;
  long long int ans = 0;
  for (long long int i = 0; i <= x; i++) {
    for (long long int j = 1; j <= min(c, k); j++) {
      ans = (ans + (dp[i][j] * ((c - j + 1 + MOD) % MOD)) % MOD) % MOD;
    }
  }
  for (long long int i = x + 1; i < n; i++) {
    for (long long int j = 1; j <= min(c - 1, k); j++) {
      ans = (ans + (dp[i][j] * ((c - j + MOD) % MOD)) % MOD) % MOD;
    }
  }
  cout << ans << '\n';
}
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}

```