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 std::cout;
using std::endl;
using std::sort;
using std::vector;
long long const MOD = 1000000007;
long long GN() {
  long long ret = 0;
  char c;
  while ((c = getchar()) < '0' || '9' < c) {
  }
  ret = c - '0';
  while ('0' <= (c = getchar()) && c <= '9') {
    ret = ret * 10 + c - '0';
  }
  return ret;
}
long long Solve(long long n, long long l, long long k, vector<long long> a) {
  vector<int> id(n);
  for (int i = 0; i < n; ++i) id[i] = i;
  sort(id.begin(), id.end(), [&a](int i, int j) { return a[i] < a[j]; });
  vector<long long> ct;
  vector<long long> inr;
  for (int i = 0, j = 0; i < n; i = j) {
    inr.push_back(0);
    while (j < n && a[id[i]] == a[id[j]]) {
      if (id[j] < l % n) {
        inr.back() += 1;
      }
      ++j;
    }
    ct.push_back(j - i);
  }
  int n2 = ct.size();
  long long ans = 0;
  long long fill = l / n;
  vector<long long> dp_now(n2);
  vector<long long> dp_sum(n2, 1);
  for (int kk = 1; kk <= k; ++kk) {
    if (fill >= kk - 1) {
      for (int nn = 0; nn < n2; ++nn) {
        ans += inr[nn] * dp_sum[nn] % MOD;
        ans %= MOD;
      }
    }
    for (int nn = 0; nn < n2; ++nn) {
      dp_now[nn] = dp_sum[nn] * ct[nn] % MOD;
    }
    dp_sum[0] = dp_now[0];
    for (int nn = 1; nn < n2; ++nn) {
      dp_sum[nn] = (dp_sum[nn - 1] + dp_now[nn]) % MOD;
    }
    if (fill >= kk) {
      ans += dp_sum[n2 - 1] * ((fill - kk + 1) % MOD) % MOD;
      ans %= MOD;
    }
  }
  return ans;
}
int main() {
  long long n, l, k;
  n = GN();
  l = GN();
  k = GN();
  vector<long long> a(n);
  for (int i = 0; i < n; ++i) {
    a[i] = GN();
  }
  cout << Solve(n, l, k, a) << endl;
  return 0;
}
