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 modpow(long long x, long long p, long long M) {
  long long ret = 1;
  while (p) {
    if (p & 1) ret = ret * x % M;
    x = x * x % M;
    p /= 2;
  }
  return ret;
}
const long long MN = 1000010;
long long n, l, k;
long long a[MN], b[MN];
long long dp[MN], cnt[MN];
map<long long, long long> m;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) cin >> a[i], m[a[i]]++;
  int len = 0;
  for (auto it = m.begin(); it != m.end(); it++, len++) {
    b[len] = dp[len] = cnt[len] = it->second;
    it->second = len;
  }
  long long ans = 0, sum, tmp;
  long long N = l / n, muln = N % 1000000007;
  for (int i = 0; i < len; i++)
    ans = (ans + muln * dp[i] % 1000000007) % 1000000007;
  for (long long ll = 2; ll <= k && ll <= N; ll++) {
    sum = 0;
    for (int i = 0; i < len; i++) sum = (sum + dp[i]) % 1000000007;
    for (int i = 0; i < len; i++) {
      tmp = dp[i];
      dp[i] = sum * cnt[i] % 1000000007;
      sum = (sum - tmp + 1000000007) % 1000000007;
      ans = (ans + (muln - ll + 1 + 1000000007) * dp[i] % 1000000007) %
            1000000007;
    }
  }
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < (l % n); i++) dp[m[a[i]]]++;
  for (int i = 0; i < len; i++) ans = (ans + dp[i]) % 1000000007;
  N += ((l % n) != 0);
  for (long long ll = 2; ll <= k && ll <= N; ll++) {
    sum = 0;
    for (int i = 0; i < len; i++) sum = (sum + dp[i]) % 1000000007;
    for (int i = 0; i < len; i++) {
      tmp = dp[i];
      dp[i] = sum * cnt[i] % 1000000007;
      sum = (sum - tmp + 1000000007) % 1000000007;
      ans = (ans + dp[i]) % 1000000007;
    }
  }
  cout << ans << "\n";
  return 0;
}
