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;
vector<long long> dp[1111111];
long long num[1111111];
long long a[1111111];
int main() {
  long long n, l, k, i, j;
  scanf("%I64d %I64d %I64d", &n, &l, &k);
  for (i = 1; i <= n; i++) scanf("%I64d", &num[i]);
  for (i = 1; i <= n; i++) a[i] = num[i];
  sort(&num[1], &num[1] + n);
  long long mod = l % n;
  long long temp = l / n;
  long long f = k;
  k = min(k, temp);
  long long res = 0;
  for (i = 0; i <= k + 2; i++) {
    dp[i].reserve(n + 2);
  }
  dp[1][0] = 0;
  for (i = 1; i <= n; i++) {
    dp[1][i] = i;
    res += temp;
    res %= 1000000007;
  }
  long long sum = 0;
  for (i = 2; i <= k; i++) {
    dp[i][0] = 0;
    int start = 1;
    j = 1;
    while (j <= n) {
      if (num[j] == num[start])
        j++;
      else {
        for (int kk = start; kk < j; kk++) {
          dp[i][kk] = dp[i - 1][j - 1] % 1000000007;
          res += (temp - i + 1) % 1000000007 * dp[i][kk];
          res %= 1000000007;
          dp[i][kk] += dp[i][kk - 1];
          dp[i][kk] %= 1000000007;
        }
        start = j;
      }
    }
    for (int kk = start; kk <= n; kk++) {
      dp[i][kk] = dp[i - 1][n] % 1000000007;
      res += (temp - i + 1) % 1000000007 * dp[i][kk];
      res %= 1000000007;
      dp[i][kk] += dp[i][kk - 1];
      dp[i][kk] %= 1000000007;
    }
  }
  sort(&a[1], &a[1] + mod);
  j = 1;
  res += mod;
  if (l / n < f) k++;
  for (i = 1; i <= mod; i++) {
    while (a[i] >= num[j] && j <= n) j++;
    for (int kk = 1; kk < k; kk++) {
      res += dp[kk][j - 1];
      res %= 1000000007;
    }
  }
  printf("%I64d\n", res % 1000000007);
  return 0;
}
