#include <bits/stdc++.h>
using namespace std;
long long modpow(long long a, long long k, long long p) {
  long long ret = 1;
  while (k) {
    if (k % 2) ret = (ret * a) % p;
    k /= 2;
    a = (a * a) % p;
  }
  return ret;
}
int main() {
  long long n, l, k;
  cin >> n >> l >> k;
  vector<int> a(n);
  map<int, int> nums, rem;
  for (__typeof(n) i = 0; i < n; ++i) {
    scanf("%d", &(a[i]));
    nums[a[i]]++;
    if (i < l % n) rem[a[i]]++;
  }
  vector<int> num, cnt, cum;
  long long s = 0;
  for (auto& it : nums) {
    num.push_back(it.first);
    cnt.push_back(it.second);
    s += it.second;
    cum.push_back(s);
  }
  int dist = num.size();
  vector<vector<long long>> ways(k, vector<long long>(dist));
  for (__typeof(k) len = 0; len < k; ++len) {
    ways[len][0] = modpow(cnt[0], len + 1, 1000000007);
  }
  for (__typeof(dist) min = 1; min < dist; ++min) {
    ways[0][min] = cum[min];
    for (__typeof(k) len = 1; len < k; ++len) {
      ways[len][min] =
          (ways[len][min - 1] + cnt[min] * ways[len - 1][min] % 1000000007) %
          1000000007;
    }
  }
  s = 0;
  for (__typeof(((l / n) > (k) ? (k) : (l / n))) i = 0;
       i < ((l / n) > (k) ? (k) : (l / n)); ++i) {
    s = (s + ((l / n - i) % 1000000007) * ways[i][dist - 1] % 1000000007) %
        1000000007;
  }
  for (auto& it : rem) {
    int pos = lower_bound(num.begin(), num.end(), it.first) - num.begin();
    s = (s + it.second) % 1000000007;
    for (__typeof(((l / n) > (k - 1) ? (k - 1) : (l / n))) i = 0;
         i < ((l / n) > (k - 1) ? (k - 1) : (l / n)); ++i) {
      s = (s + it.second * ways[i][pos] % 1000000007) % 1000000007;
    }
  }
  cout << s << endl;
  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;
long long modpow(long long a, long long k, long long p) {
  long long ret = 1;
  while (k) {
    if (k % 2) ret = (ret * a) % p;
    k /= 2;
    a = (a * a) % p;
  }
  return ret;
}
int main() {
  long long n, l, k;
  cin >> n >> l >> k;
  vector<int> a(n);
  map<int, int> nums, rem;
  for (__typeof(n) i = 0; i < n; ++i) {
    scanf("%d", &(a[i]));
    nums[a[i]]++;
    if (i < l % n) rem[a[i]]++;
  }
  vector<int> num, cnt, cum;
  long long s = 0;
  for (auto& it : nums) {
    num.push_back(it.first);
    cnt.push_back(it.second);
    s += it.second;
    cum.push_back(s);
  }
  int dist = num.size();
  vector<vector<long long>> ways(k, vector<long long>(dist));
  for (__typeof(k) len = 0; len < k; ++len) {
    ways[len][0] = modpow(cnt[0], len + 1, 1000000007);
  }
  for (__typeof(dist) min = 1; min < dist; ++min) {
    ways[0][min] = cum[min];
    for (__typeof(k) len = 1; len < k; ++len) {
      ways[len][min] =
          (ways[len][min - 1] + cnt[min] * ways[len - 1][min] % 1000000007) %
          1000000007;
    }
  }
  s = 0;
  for (__typeof(((l / n) > (k) ? (k) : (l / n))) i = 0;
       i < ((l / n) > (k) ? (k) : (l / n)); ++i) {
    s = (s + ((l / n - i) % 1000000007) * ways[i][dist - 1] % 1000000007) %
        1000000007;
  }
  for (auto& it : rem) {
    int pos = lower_bound(num.begin(), num.end(), it.first) - num.begin();
    s = (s + it.second) % 1000000007;
    for (__typeof(((l / n) > (k - 1) ? (k - 1) : (l / n))) i = 0;
         i < ((l / n) > (k - 1) ? (k - 1) : (l / n)); ++i) {
      s = (s + it.second * ways[i][pos] % 1000000007) % 1000000007;
    }
  }
  cout << s << endl;
  return 0;
}

```