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>.
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e+9 + 7;
bool cmp(const pair<pair<int, int>, int>& p1,
         const pair<pair<int, int>, int>& p2) {
  if (p1.second != p2.second) return p1.second < p2.second;
  return p1.first < p2.first;
}
int main() {
  int n, k;
  long long l;
  cin >> n >> l >> k;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<int> dp(n * k);
  vector<pair<pair<int, int>, int>> s(n);
  for (int i = 0; i < n; i++) {
    s[i] = make_pair(make_pair(a[i], 0), i);
  }
  vector<int> sum(n + 1);
  for (int i = 0; i < k; i++) {
    sort(s.begin(), s.end());
    if (i != 0) {
      for (int j = 0; j < n; j++) {
        auto it = lower_bound(s.begin(), s.end(),
                              make_pair(make_pair(s[j].first.first + 1, 0), 0));
        int cnt = distance(s.begin(), it);
        s[j].first.second = sum[cnt];
      }
    } else {
      for (int j = 1; j < n + 1; j++) {
        s[j - 1].first.second = 1;
      }
    }
    for (int i = 1; i <= n; i++) {
      sum[i] = (sum[i - 1] + s[i - 1].first.second) % MOD;
    }
    sort(s.begin(), s.end(), cmp);
    for (int j = 0; j < n; j++) {
      dp[i * n + j] = s[j].first.second;
    }
  }
  int ans = 0;
  for (int i = 0; i < k; i++) {
    int s1 = 0;
    for (int j = i * n; j < (i + 1) * n; j++) s1 = (s1 + dp[j]) % MOD;
    if (l >= i * n) {
      ans = (ans + (((l - i * n) / n) % MOD) * 1ll * s1) % MOD;
      s1 = 0;
      for (int j = 0; j < l % n; j++) s1 = (s1 + dp[i * n + j]) % MOD;
      ans = (ans + s1) % MOD;
    }
  }
  cout << ans;
  return 0;
}
