#include <bits/stdc++.h>
using namespace std;
const long double pi = 3.1415926535897932384626433832795l;
template <typename T>
inline auto sqr(T x) -> decltype(x * x) {
  return x * x;
}
template <typename T1, typename T2>
inline bool umx(T1& a, T2 b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T1, typename T2>
inline bool umn(T1& a, T2 b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
const int64_t mod = 1000000007;
const int N = 1000000;
struct Input {
  int n, k;
  int64_t l;
  int a[N];
  bool read() {
    if (!(cin >> n >> l >> k)) {
      return 0;
    }
    for (int i = int(0); i < int(n); ++i) {
      scanf("%d", &a[i]);
    }
    return 1;
  }
  void init(const Input& input) { *this = input; }
};
struct Data : Input {
  int64_t ans;
  void write() { cout << ans << endl; }
  virtual void solve() {}
  virtual void clear() { *this = Data(); }
};
struct Solution : Data {
  int64_t pw(int64_t a, int64_t n) {
    int64_t ans = 1;
    while (n) {
      if (n & 1) {
        (ans *= a) %= mod;
      }
      (a *= a) %= mod;
      n >>= 1;
    }
    return ans;
  }
  int64_t d[2][N];
  void solve() {
    int64_t q = l / n;
    int r = l % n;
    vector<pair<int, int> > v;
    for (int i = int(0); i < int(n); ++i) {
      v.push_back(make_pair(a[i], i));
    }
    sort((v).begin(), (v).end());
    memset(d, 0, sizeof d);
    for (int i = int(0); i < int(n); ++i) {
      d[1][i] = 1;
    }
    for (int x = int(1); x < int(k + 1); ++x) {
      int64_t cq = max<int64_t>(0, q - x + 1) % mod;
      int64_t cq1 = max<int64_t>(0, q - x + 2) % mod;
      int x0 = x & 1;
      int x1 = x0 ^ 1;
      for (int i = int(0); i < int(n); ++i) {
        ;
        (ans += d[x0][i] * ((i < r) ? cq1 : cq)) %= mod;
      }
      int64_t s = 0;
      for (int l = 0; l < n;) {
        int r = l;
        while (r < n && v[r].first == v[l].first) {
          (s += d[x0][v[r].second]) %= mod;
          ++r;
        }
        for (int i = int(l); i < int(r); ++i) {
          d[x1][v[i].second] = s;
        }
        l = r;
      }
    }
  }
  void clear() { *this = Solution(); }
};
Solution sol;
int main() {
  cout.setf(ios::showpoint | ios::fixed);
  cout.precision(20);
  sol.read();
  sol.solve();
  sol.write();
  return 0;
}

### Prompt

Your challenge is to write 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 double pi = 3.1415926535897932384626433832795l;
template <typename T>
inline auto sqr(T x) -> decltype(x * x) {
  return x * x;
}
template <typename T1, typename T2>
inline bool umx(T1& a, T2 b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T1, typename T2>
inline bool umn(T1& a, T2 b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
const int64_t mod = 1000000007;
const int N = 1000000;
struct Input {
  int n, k;
  int64_t l;
  int a[N];
  bool read() {
    if (!(cin >> n >> l >> k)) {
      return 0;
    }
    for (int i = int(0); i < int(n); ++i) {
      scanf("%d", &a[i]);
    }
    return 1;
  }
  void init(const Input& input) { *this = input; }
};
struct Data : Input {
  int64_t ans;
  void write() { cout << ans << endl; }
  virtual void solve() {}
  virtual void clear() { *this = Data(); }
};
struct Solution : Data {
  int64_t pw(int64_t a, int64_t n) {
    int64_t ans = 1;
    while (n) {
      if (n & 1) {
        (ans *= a) %= mod;
      }
      (a *= a) %= mod;
      n >>= 1;
    }
    return ans;
  }
  int64_t d[2][N];
  void solve() {
    int64_t q = l / n;
    int r = l % n;
    vector<pair<int, int> > v;
    for (int i = int(0); i < int(n); ++i) {
      v.push_back(make_pair(a[i], i));
    }
    sort((v).begin(), (v).end());
    memset(d, 0, sizeof d);
    for (int i = int(0); i < int(n); ++i) {
      d[1][i] = 1;
    }
    for (int x = int(1); x < int(k + 1); ++x) {
      int64_t cq = max<int64_t>(0, q - x + 1) % mod;
      int64_t cq1 = max<int64_t>(0, q - x + 2) % mod;
      int x0 = x & 1;
      int x1 = x0 ^ 1;
      for (int i = int(0); i < int(n); ++i) {
        ;
        (ans += d[x0][i] * ((i < r) ? cq1 : cq)) %= mod;
      }
      int64_t s = 0;
      for (int l = 0; l < n;) {
        int r = l;
        while (r < n && v[r].first == v[l].first) {
          (s += d[x0][v[r].second]) %= mod;
          ++r;
        }
        for (int i = int(l); i < int(r); ++i) {
          d[x1][v[i].second] = s;
        }
        l = r;
      }
    }
  }
  void clear() { *this = Solution(); }
};
Solution sol;
int main() {
  cout.setf(ios::showpoint | ios::fixed);
  cout.precision(20);
  sol.read();
  sol.solve();
  sol.write();
  return 0;
}

```