#include <bits/stdc++.h>
using namespace std;
template <class T, class U>
ostream& operator<<(ostream& o, const pair<T, U>& v) {
  return o << "(" << v.first << ", " << v.second << ")";
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& v) {
  o << "{";
  for (int i = 0; i < (int)((int)((v).size())); ++i)
    o << (i ? ", " : "") << v[i];
  return o << "}";
}
template <class T, class U>
ostream& operator<<(ostream& o, const map<T, U>& v) {
  o << "{";
  for (const auto& e : v) o << e << ", ";
  return o << "}";
}
template <class T, class U>
ostream& operator<<(ostream& o, const unordered_map<T, U>& v) {
  o << "{";
  for (const auto& e : v) o << e << ", ";
  return o << "}";
}
template <class T>
ostream& operator<<(ostream& o, const set<T>& v) {
  o << "{";
  for (const auto& e : v) o << e << ", ";
  return o << "}";
}
template <class T>
string to_s(const T& v) {
  ostringstream is;
  is << v;
  return is.str();
}
const int dx[] = {0, 1, 0, -1, 1, 1, -1, -1, 0};
const int dy[] = {1, 0, -1, 0, -1, 1, 1, -1, 0};
template <class T = long long>
struct BIT {
  vector<T> bit;
  const int n;
  BIT(int N) : bit(N), n(N){};
  void add(int i, T v) {
    for (; i < n; i |= i + 1) bit[i] += v;
  }
  T sum(int i) {
    T res = T();
    --i;
    for (; i >= 0; i = (i & (i + 1)) - 1) res += bit[i];
    return res;
  }
  T sum(int a, int b) { return sum(b) - sum(a); }
  T get(int a) { return sum(a + 1) - sum(a); }
  void setv(int i, T v) { add(i, v - get(i)); }
};
const long long mod = 1e9 + 7;
long long app[1100000];
int main(int argc, char* argv[]) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, l, k;
  cin >> n >> l >> k;
  vector<int> in(n);
  vector<int> app(n);
  for (int i = 0; i < (int)(n); ++i) {
    cin >> in[i];
    app[i] = in[i];
  }
  sort((app).begin(), (app).end());
  app.erase(unique((app).begin(), (app).end()), app.end());
  for (int i = 0; i < (int)(n); ++i) {
    in[i] = lower_bound((app).begin(), (app).end(), in[i]) - app.begin() + 1;
  }
  long long ans = 0;
  BIT<> bit((int)((app).size()) + 10);
  bit.add(0, 1);
  for (int iik = 0; iik < (int)(k); ++iik) {
    BIT<> nbit((int)((app).size()) + 10);
    for (int i = 0; i < n; ++i) {
      long long t = bit.sum(in[i] + 1) % mod;
      long long le = l / n;
      if (i < l % n) ++le;
      if (le >= iik) ans += (le - iik) % mod * t % mod;
      t = (nbit.get(in[i]) + t) % mod;
      nbit.setv(in[i], t);
    }
    nbit.bit.swap(bit.bit);
    ans %= mod;
  }
  cout << ans << '\n';
}

### 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;
template <class T, class U>
ostream& operator<<(ostream& o, const pair<T, U>& v) {
  return o << "(" << v.first << ", " << v.second << ")";
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& v) {
  o << "{";
  for (int i = 0; i < (int)((int)((v).size())); ++i)
    o << (i ? ", " : "") << v[i];
  return o << "}";
}
template <class T, class U>
ostream& operator<<(ostream& o, const map<T, U>& v) {
  o << "{";
  for (const auto& e : v) o << e << ", ";
  return o << "}";
}
template <class T, class U>
ostream& operator<<(ostream& o, const unordered_map<T, U>& v) {
  o << "{";
  for (const auto& e : v) o << e << ", ";
  return o << "}";
}
template <class T>
ostream& operator<<(ostream& o, const set<T>& v) {
  o << "{";
  for (const auto& e : v) o << e << ", ";
  return o << "}";
}
template <class T>
string to_s(const T& v) {
  ostringstream is;
  is << v;
  return is.str();
}
const int dx[] = {0, 1, 0, -1, 1, 1, -1, -1, 0};
const int dy[] = {1, 0, -1, 0, -1, 1, 1, -1, 0};
template <class T = long long>
struct BIT {
  vector<T> bit;
  const int n;
  BIT(int N) : bit(N), n(N){};
  void add(int i, T v) {
    for (; i < n; i |= i + 1) bit[i] += v;
  }
  T sum(int i) {
    T res = T();
    --i;
    for (; i >= 0; i = (i & (i + 1)) - 1) res += bit[i];
    return res;
  }
  T sum(int a, int b) { return sum(b) - sum(a); }
  T get(int a) { return sum(a + 1) - sum(a); }
  void setv(int i, T v) { add(i, v - get(i)); }
};
const long long mod = 1e9 + 7;
long long app[1100000];
int main(int argc, char* argv[]) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, l, k;
  cin >> n >> l >> k;
  vector<int> in(n);
  vector<int> app(n);
  for (int i = 0; i < (int)(n); ++i) {
    cin >> in[i];
    app[i] = in[i];
  }
  sort((app).begin(), (app).end());
  app.erase(unique((app).begin(), (app).end()), app.end());
  for (int i = 0; i < (int)(n); ++i) {
    in[i] = lower_bound((app).begin(), (app).end(), in[i]) - app.begin() + 1;
  }
  long long ans = 0;
  BIT<> bit((int)((app).size()) + 10);
  bit.add(0, 1);
  for (int iik = 0; iik < (int)(k); ++iik) {
    BIT<> nbit((int)((app).size()) + 10);
    for (int i = 0; i < n; ++i) {
      long long t = bit.sum(in[i] + 1) % mod;
      long long le = l / n;
      if (i < l % n) ++le;
      if (le >= iik) ans += (le - iik) % mod * t % mod;
      t = (nbit.get(in[i]) + t) % mod;
      nbit.setv(in[i], t);
    }
    nbit.bit.swap(bit.bit);
    ans %= mod;
  }
  cout << ans << '\n';
}

```