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>
#pragma comment(linker, "/STACK:100000000000000")
using namespace std;
long long n, k, l, arr[1000011];
int startPos[1000011], startPos2[1000011];
vector<long long> dp[1000011], vec, vec2;
long long rec(int p, int rem) {
  if (rem == 0) return 1;
  if (p >= vec.size()) return 0;
  if (~dp[p][rem]) return dp[p][rem];
  long long ret = rec(startPos[p], rem - 1) + rec(p + 1, rem);
  if (ret >= 1000000007) ret %= 1000000007;
  return dp[p][rem] = ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> l >> k;
  vector<long long> em(k + 1, -1);
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    vec.push_back(arr[i]);
    dp[i] = em;
  }
  sort((vec).begin(), (vec).end());
  reverse((vec).begin(), (vec).end());
  int p = 0;
  for (int i = 0; i < vec.size(); i++)
    if (vec[p] == vec[i]) {
      startPos[i] = p;
    } else {
      startPos[i] = i;
      p = i;
    }
  long long ans = 0, bkt = l / n;
  int mini = min(bkt, k);
  for (int j = 0; j < mini; j++)
    for (int i = 0; i < vec.size(); i++) {
      long long tmp = rec(startPos[i], j);
      long long U = (bkt - j) % 1000000007;
      (tmp *= U) %= 1000000007;
      (ans += tmp) %= 1000000007;
    }
  long long rem = l % n;
  for (int i = 0; i < rem; i++) vec2.push_back(arr[i]);
  sort((vec2).begin(), (vec2).end());
  reverse((vec2).begin(), (vec2).end());
  p = 0;
  for (int i = 0; i < vec2.size(); i++) {
    while (p < vec.size() && vec[p] > vec2[i]) p++;
    startPos2[i] = p;
  }
  mini = min((l + n - 1) / n, k);
  long long bkt2 = (l + n - 1) / n;
  for (int j = 0; j < mini; j++)
    for (int i = 0; i < vec2.size(); i++) {
      long long tmp = rec(startPos2[i], j);
      (ans += tmp) %= 1000000007;
    }
  cout << ans << endl;
  return 0;
}
