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 MAXN = 1000001;
const long long MOD = 1e9 + 7;
struct Tree {
  int n;
  long long a[MAXN];
  long long t[MAXN];
  Tree(int n = 0) : n(n) {
    for (int i = 0; i < n; i++) a[i] = 0;
    for (int i = 0; i < n; i++) t[i] = 0;
  }
  void init(int n1 = 0) {
    n = n1;
    for (int i = 0; i < n; i++) a[i] = 0;
    for (int i = 0; i < n; i++) t[i] = 0;
  }
  void add(int p, int d) {
    for (; p < n; p = p | (p + 1)) t[p] = (t[p] + d) % MOD;
  }
  long long get(int k) {
    long long res = 0;
    for (; k >= 0; k = (k & (k + 1)) - 1) res = (res + t[k]) % MOD;
    return res;
  }
  void print() {
    for (int i = 0; i < n; i++) cout << get(i) - get(i - 1) << " ";
    cout << endl;
  }
};
int a[MAXN];
Tree tr1, tr2;
int main() {
  ios_base::sync_with_stdio(false);
  long long n, l, k;
  cin >> n >> l >> k;
  vector<pair<int, int> > x;
  x.reserve(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    x.push_back(make_pair(a[i], i));
  }
  sort(x.begin(), x.end());
  a[x[0].second] = 0;
  for (int i = 1; i < n; i++) {
    if (x[i].first != x[i - 1].first)
      a[x[i].second] = a[x[i - 1].second] + 1;
    else
      a[x[i].second] = a[x[i - 1].second];
  }
  tr1.init(n);
  tr2.init(n);
  tr2.add(0, 1);
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    for (int j = 0; j < n; j++)
      tr1.a[j] = tr2.a[j], tr1.t[j] = tr2.t[j], tr2.a[j] = tr2.t[j] = 0;
    for (int j = 0; j < n; j++) {
      long long p = tr1.get(a[j]);
      tr2.add(a[j], p);
      if (i <= l / n)
        ans = (ans +
               ((l / n) - i + (long long)((i <= l / n) && (j < l % n)) % MOD) %
                   MOD * p) %
              MOD;
    }
  }
  cout << ans << endl;
  return 0;
}
