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 A[1000005], B[1000005], cnt[1000005], num[1000005];
long long dp[2][1000005];
int main() {
  int n, k;
  long long l, ans = 0;
  cin >> n >> l >> k;
  if (l <= n) {
    cout << l << endl;
    return 0;
  }
  for (int i = 0; i < n; i++) scanf("%d", &A[i]);
  for (int i = 0; i < n; i++) B[i] = A[i];
  sort(B, B + n);
  int m = 0, a = 0;
  long long S = l / n;
  for (int i = 0; i < n; i++) {
    a++;
    if (B[i] != B[i + 1] || i == n - 1) {
      B[m] = B[i];
      cnt[m++] = a;
      a = 0;
    }
  }
  sort(A, A + l % n);
  for (int i = 0; i < m; i++) {
    int t = lower_bound(A, A + l % n, B[i]) - A;
    num[i] = l % n - t;
  }
  dp[0][0] = 1;
  for (int i = 1; i <= min(S, 1LL * k); i++) {
    int cur = i & 1;
    long long sum = 0;
    for (int j = 0; j < n; j++) {
      sum = (sum + dp[!cur][j]) % 1000000007;
      dp[cur][j] = (sum * cnt[j]) % 1000000007;
      ans = (ans + dp[cur][j] * (S % 1000000007 + 1000000007 - i + 1)) %
            1000000007;
      if (i < k) ans = (ans + dp[cur][j] * num[j]) % 1000000007;
    }
  }
  ans = (ans + l % n) % 1000000007;
  cout << ans << endl;
  return 0;
}
