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];
int dp[2][1000005];
int sum[2][1000005];
long long n, l, k;
int bs(int x) {
  int l = 0, r = n;
  while (l < r - 1) {
    int mid = (l + r) >> 1;
    if (b[mid] <= x)
      l = mid;
    else
      r = mid;
  }
  return l;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  while (cin >> n >> l >> k) {
    for (int i = (0); i <= (n - 1); i++) {
      cin >> a[i];
      b[i] = a[i];
    }
    sort(b, b + n);
    memset(dp, 0, sizeof(dp));
    memset(sum, 0, sizeof(sum));
    long long num = l / n + 1;
    long long ret = l % n;
    k = min(num, k);
    for (int i = (0); i <= (n); i++) sum[0][i] = 1;
    long long ans = 0;
    for (int i = (1); i <= (k); i++) {
      int f = i & 1;
      for (int j = (0); j <= (n - 1); j++) {
        sum[f][j] = dp[f][j] = 0;
      }
      for (int j = (0); j <= (n - 1); j++) {
        int p = bs(a[j]);
        dp[f][j] = sum[1 - f][p];
        sum[f][p] = (sum[f][p] + dp[f][j]) % 1000000007;
        if (j >= ret)
          ans = (ans + (num - i) % 1000000007 * dp[f][j] % 1000000007) %
                1000000007;
        else
          ans = (ans + (num - i + 1) % 1000000007 * dp[f][j] % 1000000007) %
                1000000007;
      }
      for (int j = (1); j <= (n - 1); j++)
        sum[f][j] = (sum[f][j] + sum[f][j - 1]) % 1000000007;
    }
    cout << ans << endl;
  }
  return 0;
}
