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;
int dp[2][1000001];
vector<pair<int, int> > a(1000001);
int main() {
  int i, j, k, n;
  long long l;
  cin >> n >> l >> k;
  int kk = k;
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i].first);
    a[i].second = i;
  }
  sort(a.begin(), a.begin() + n);
  long long ans = 0;
  long long lev = l / n;
  int r = l % n - 1;
  for (i = 0; i < n; i++) {
    dp[0][i] = 1;
    ans = (ans + lev);
    if (a[i].second <= r) ans++;
    if (ans >= 1000000007) ans %= 1000000007;
  }
  int p = 0;
  long long maxlen = lev;
  if (r > -1) maxlen++;
  for (j = 2; j <= min((long long)k, maxlen); j++) {
    if (j > kk) exit(-1);
    p = 1 - p;
    int loc = 0;
    long long sum = 0;
    for (i = 0; i < n; i++) {
      while (loc < n && a[loc].first <= a[i].first)
        sum = (sum + dp[!p][loc++]) % 1000000007;
      dp[p][i] = sum;
      ans = (ans + sum * ((lev - j + 1) % 1000000007)) % 1000000007;
      if (a[i].second <= r) ans = (ans + sum) % 1000000007;
    }
  }
  cout << ans % 1000000007 << endl;
  return 0;
}
