#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-9;
const int inf = 2000000000;
const long long infLL = 9000000000000000000;
template <typename first, typename second>
ostream& operator<<(ostream& os, const pair<first, second>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ",";
    os << *it;
  }
  return os << "]";
}
template <typename T>
ostream& operator<<(ostream& os, const multiset<T>& v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename first, typename second>
ostream& operator<<(ostream& os, const map<first, second>& v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
void faltu() { cerr << '\n'; }
template <typename T>
void faltu(T a[], int n) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << '\n';
}
template <typename T, typename... hello>
void faltu(T arg, const hello&... rest) {
  cerr << arg << ' ';
  faltu(rest...);
}
long long n, l, k;
vector<long long> a;
vector<long long> store;
vector<vector<long long>> dp;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> l >> k;
  dp.resize(n + 2, vector<long long>(k + 2));
  a.resize(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    store.push_back(a[i]);
  }
  sort(store.begin(), store.end());
  for (int i = 0; i < n; ++i) {
    a[i] = (lower_bound(store.begin(), store.end(), a[i]) - store.begin()) + 1;
  }
  long long ans = 0;
  for (int i = 0; i < min(n, l); ++i) {
    dp[a[i]][1] += 1;
    if (l - 1 - i < 0) continue;
    ans += (l - 1 - i) / n + 1;
    ans %= 1000000007;
  }
  for (int i = 1; i <= n; ++i) {
    dp[i][1] += dp[i - 1][1];
    dp[i][1] %= 1000000007;
  }
  for (int i = 2; i <= min(k, l); ++i) {
    for (int j = 0; j < min(n, l); ++j) {
      if (l - 1 - (j + (i - 1) * n) < 0) continue;
      long long temp = dp[a[j]][i - 1] *
                       (((l - 1 - (j + (i - 1) * n)) / n + 1) % 1000000007);
      temp %= 1000000007;
      dp[a[j]][i] += dp[a[j]][i - 1];
      dp[a[j]][i] %= 1000000007;
      ans = (ans + temp) % 1000000007;
    }
    for (int j = 1; j <= n; ++j) {
      dp[j][i] += dp[j - 1][i];
      dp[j][i] %= 1000000007;
    }
  }
  cout << ans << '\n';
}

### Prompt

In CPP, your task is to solve 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 double eps = 1e-9;
const int inf = 2000000000;
const long long infLL = 9000000000000000000;
template <typename first, typename second>
ostream& operator<<(ostream& os, const pair<first, second>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ",";
    os << *it;
  }
  return os << "]";
}
template <typename T>
ostream& operator<<(ostream& os, const multiset<T>& v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename first, typename second>
ostream& operator<<(ostream& os, const map<first, second>& v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
void faltu() { cerr << '\n'; }
template <typename T>
void faltu(T a[], int n) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << '\n';
}
template <typename T, typename... hello>
void faltu(T arg, const hello&... rest) {
  cerr << arg << ' ';
  faltu(rest...);
}
long long n, l, k;
vector<long long> a;
vector<long long> store;
vector<vector<long long>> dp;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> l >> k;
  dp.resize(n + 2, vector<long long>(k + 2));
  a.resize(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    store.push_back(a[i]);
  }
  sort(store.begin(), store.end());
  for (int i = 0; i < n; ++i) {
    a[i] = (lower_bound(store.begin(), store.end(), a[i]) - store.begin()) + 1;
  }
  long long ans = 0;
  for (int i = 0; i < min(n, l); ++i) {
    dp[a[i]][1] += 1;
    if (l - 1 - i < 0) continue;
    ans += (l - 1 - i) / n + 1;
    ans %= 1000000007;
  }
  for (int i = 1; i <= n; ++i) {
    dp[i][1] += dp[i - 1][1];
    dp[i][1] %= 1000000007;
  }
  for (int i = 2; i <= min(k, l); ++i) {
    for (int j = 0; j < min(n, l); ++j) {
      if (l - 1 - (j + (i - 1) * n) < 0) continue;
      long long temp = dp[a[j]][i - 1] *
                       (((l - 1 - (j + (i - 1) * n)) / n + 1) % 1000000007);
      temp %= 1000000007;
      dp[a[j]][i] += dp[a[j]][i - 1];
      dp[a[j]][i] %= 1000000007;
      ans = (ans + temp) % 1000000007;
    }
    for (int j = 1; j <= n; ++j) {
      dp[j][i] += dp[j - 1][i];
      dp[j][i] %= 1000000007;
    }
  }
  cout << ans << '\n';
}

```