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 = 1e6 + 100;
const int M = 2e6 + 100;
const int sq = 100;
const long long int mod = 1e9 + 7;
const int inf = 1e9;
long long int a[N], b[N], dp[M];
vector<pair<int, int>> v;
long long int n, l, k;
inline int id(int x, int y) {
  if (k + 1 > n) return (x * (k + 1) + y);
  return (y * (n + 1) + x);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> l >> k;
  long long int l1 = l / n, r = l % n, ans = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(a, a + n);
  for (int i = 0; i < n;) {
    int j = i;
    while (i < n && a[i] == a[j]) i++;
    v.push_back({a[j], i - j});
  }
  n = v.size();
  for (int i = 0; i < n; i++) {
    int cnt = v[i].second;
    dp[id(i, 0)] = 1;
    for (int j = 1; j <= k; j++) {
      dp[id(i, j)] = (dp[id(i, j - 1)] * cnt) % mod;
      if (i != 0) dp[id(i, j)] += dp[id(i - 1, j)];
      dp[id(i, j)] %= mod;
    }
  }
  for (int x = 1; x <= min(k, l1 + (r != 0)); x++) {
    long long int res = (((l1 - x + 1) % mod) * dp[id(n - 1, x)]) % mod;
    for (int i = 0; i < r; i++) {
      pair<int, int> p = make_pair(b[i], -1);
      int pt = lower_bound(v.begin(), v.end(), p) - v.begin();
      res = (res + dp[id(pt, x - 1)]) % mod;
    }
    ans = (ans + res) % mod;
  }
  cout << ans << endl;
  return 0;
}
