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 INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
const int MOD = 1e9 + 7, MAX = 1e6 + 10;
long long add(long long a, long long b) { return ((a + b) % MOD + MOD) % MOD; }
long long mul(long long a, long long b) {
  return ((a % MOD) * (b % MOD) % MOD + MOD) % MOD;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, l, k;
  cin >> n >> l >> k;
  vector<pair<int, int>> a(n);
  int ctr = 0;
  for (auto& [ai, i] : a) {
    cin >> ai;
    i = ctr++;
  }
  sort(a.begin(), a.end());
  vector<int> look(n);
  for (int i = n - 1; i >= 0; i--) {
    if (i == n - 1 or a[i].first != a[i + 1].first)
      look[i] = i;
    else
      look[i] = look[i + 1];
  }
  vector<long long> memo(n, 1), pref(n, 1);
  long long ans = 0;
  for (int sz = 0; sz < k; sz++) {
    for (int i = n - 1; i >= 0; i--) memo[i] = pref[look[i]];
    for (int i = 0; i < n; i++) {
      long long times = max(0LL, l / n - sz + (a[i].second < l % n));
      ans = add(ans, mul(times, memo[i]));
    }
    for (int i = 0; i < n; i++) {
      pref[i] = memo[i];
      if (i) pref[i] = add(pref[i], pref[i - 1]);
    }
  }
  cout << ans << '\n';
  return 0;
}
