#include <bits/stdc++.h>
using namespace std;
template <class T, class U>
bool cmax(T& a, const U& b) {
  return a < b ? a = b, 1 : 0;
}
template <class T, class U>
bool cmin(T& a, const U& b) {
  return b < a ? a = b, 1 : 0;
}
void _BG(const char* s) { cerr << s << endl; }
template <class T, class... TT>
void _BG(const char* s, T a, TT... b) {
  for (size_t c = 0; *s && (c || *s != ','); cerr << *s++)
    c += count(begin("([{"), end("([{"), *s) -
         count(begin(")]}"), end(")]}"), *s);
  cerr << " = " << a;
  if (*s) {
    cerr << ", ";
    ++s;
  }
  _BG(s, b...);
}
const int N = 1e6 + 87, MOD = 1e9 + 7;
pair<int, int> a[N];
long long dp[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  long long m;
  cin >> n >> m >> k;
  long long q = m / n, r = m % n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a, a + n);
  long long ans = 0;
  for (int x = 1; x <= k && x <= q + (r > 0); ++x) {
    long long cq1 = x <= q ? (q - x + 1) % MOD : 0;
    long long sum = x == 1 ? 1 : 0;
    for (int i = 0, j = 0; i < n; ++i) {
      while (j < n && a[j].first <= a[i].first) (sum += dp[j++]) %= MOD;
      dp[i] = sum;
      ans += dp[i] * cq1 % MOD;
      if (a[i].second < r) ans += dp[i];
    }
  }
  cout << ans % MOD << endl;
}

### 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;
template <class T, class U>
bool cmax(T& a, const U& b) {
  return a < b ? a = b, 1 : 0;
}
template <class T, class U>
bool cmin(T& a, const U& b) {
  return b < a ? a = b, 1 : 0;
}
void _BG(const char* s) { cerr << s << endl; }
template <class T, class... TT>
void _BG(const char* s, T a, TT... b) {
  for (size_t c = 0; *s && (c || *s != ','); cerr << *s++)
    c += count(begin("([{"), end("([{"), *s) -
         count(begin(")]}"), end(")]}"), *s);
  cerr << " = " << a;
  if (*s) {
    cerr << ", ";
    ++s;
  }
  _BG(s, b...);
}
const int N = 1e6 + 87, MOD = 1e9 + 7;
pair<int, int> a[N];
long long dp[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  long long m;
  cin >> n >> m >> k;
  long long q = m / n, r = m % n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a, a + n);
  long long ans = 0;
  for (int x = 1; x <= k && x <= q + (r > 0); ++x) {
    long long cq1 = x <= q ? (q - x + 1) % MOD : 0;
    long long sum = x == 1 ? 1 : 0;
    for (int i = 0, j = 0; i < n; ++i) {
      while (j < n && a[j].first <= a[i].first) (sum += dp[j++]) %= MOD;
      dp[i] = sum;
      ans += dp[i] * cq1 % MOD;
      if (a[i].second < r) ans += dp[i];
    }
  }
  cout << ans % MOD << endl;
}

```