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;
long long int m = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  long long int n, l, k;
  cin >> n >> l >> k;
  vector<vector<long long int> > dp(n + 1, vector<long long int>(k + 1));
  vector<long long int> a(n + 1);
  vector<long long int> pos(n + 1);
  vector<pair<long long int, long long int> > p;
  vector<long long int> mul(n + 1);
  for (int i = 0; i <= n - 1; i++) {
    cin >> a[i];
    p.push_back(make_pair(a[i], i));
  }
  sort(p.begin(), p.end());
  long long int mu = l / n, re = l % n;
  int ptr = 0;
  for (int i = 0; i <= n - 1; i++) {
    while (ptr < n - 1 && p[ptr + 1].first <= p[i].first) ptr++;
    pos[i] = ptr;
    dp[i][0] = 1;
    if (p[i].second < re)
      mul[i] = mu + 1;
    else
      mul[i] = mu;
  }
  long long int ans = 0;
  for (int j = 1; j <= k; j++) {
    for (int i = 0; i <= n - 1; i++) {
      dp[i][j] = dp[pos[i]][j - 1];
      long long int t = mul[i];
      t -= (j - 1);
      if (t < 0) t = 0;
      t %= m;
      ans = (ans + t * dp[i][j]) % m;
      if (i) dp[i][j] += dp[i - 1][j];
      dp[i][j] %= m;
    }
  }
  cout << ans;
  return 0;
}
