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;
const int N = 1123456, lgN = 22, rootN = 1123;
const double eps = 1e-11, pi = 3.1415926535;
long long cnt(long long l, long long p, long long k, long long n) {
  long long ans = (l / n);
  if (p < l % n) ++ans;
  ans -= (k - 1);
  ans %= 1000000007LL;
  return max(0LL, ans);
}
vector<pair<long long, long long>> vec;
vector<vector<long long>> dp;
int main() {
  long long n, k, l;
  scanf("%lld %lld %lld", &n, &l, &k);
  long long x;
  for (int i = 0; i < n; ++i) {
    scanf("%lld", &x);
    vec.push_back(make_pair(x, i));
  }
  sort(vec.begin(), vec.end());
  dp.resize(n);
  for (int i = 0; i < n; ++i) dp[i].resize(k + 1);
  for (int i = 0; i < n; ++i) dp[i][1] = 1;
  for (int j = 2; j <= k; ++j) {
    long long sum = 0;
    for (int i = 0; i < n; ++i) {
      int last = i;
      while (last < n && vec[i].first == vec[last].first) {
        sum += dp[last][j - 1];
        if (sum >= 1000000007LL) sum -= 1000000007LL;
        ++last;
      }
      --last;
      for (int p = i; p <= last; ++p) dp[p][j] = sum;
      i = last;
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    int p = vec[i].second;
    for (int j = 1; j <= k; ++j) {
      long long v = dp[i][j];
      long long c = cnt(l, p, j, n);
      ans = (ans + c * v) % 1000000007LL;
    }
  }
  printf("%lld\n", ans);
  return 0;
}
