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 = 2e6 + 100;
const int M = 1e9 + 7;
long long int a[maxN], where[maxN];
vector<int> tartib, nums[maxN];
vector<pair<int, int> > v;
long long int n, k, l, ans, t, remain;
long long int p[maxN], sum[maxN];
void input() {
  cin >> n >> l >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    v.push_back(make_pair(a[i], i));
  }
  sort(v.begin(), v.end());
}
void alg() {
  for (int i = 0; i < v.size(); i++) {
    tartib.push_back(v[i].second);
    int now = i + 1;
    where[v[i].second] = v[i].second;
    if (now >= v.size()) break;
    while (v[i].first == v[now].first) {
      nums[v[i].second].push_back(v[now].second);
      where[v[now].second] = v[i].second;
      now++;
      if (now >= v.size()) break;
    }
    i = now - 1;
  }
}
void init() {
  for (int i = 0; i <= n; i++) p[i] = 1;
}
int main() {
  ios::sync_with_stdio(false);
  input();
  alg();
  t = l / n;
  remain = l - t * n;
  init();
  long long int ppp = t;
  if (remain > 0) ppp++;
  for (int j = 1; j <= min(k, ppp); j++) {
    for (int i = 1; i <= n; i++) {
      ans += (((t - j + 1) % M) * p[i]) % M;
      if (i <= remain) ans += p[i];
      ans %= M;
    }
    for (int i = 0; i < tartib.size(); i++) {
      sum[tartib[i]] =
          (long long)(nums[tartib[i]].size() + 1) * p[tartib[i]] % M;
      sum[tartib[i]] %= M;
      if (i > 0) sum[tartib[i]] += sum[tartib[i - 1]];
      sum[tartib[i]] %= M;
    }
    for (int i = 1; i <= n; i++) {
      p[i] = sum[where[i]];
    }
  }
  cout << ans << endl;
  return 0;
}
