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;
int main() {
  long long int m = 1000000007;
  int N, K;
  long long int L;
  cin >> N >> L >> K;
  vector<vector<long long int> > dp(K + 1);
  for (int i = 0; i < K + 1; i++) dp[i].resize(N);
  vector<pair<int, int> > v(N);
  for (int i = 0; i < N; i++) {
    scanf("%d", &v[i].first);
    v[i].second = i;
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < N; i++) dp[1][i] = 1;
  for (int i = 2; i <= K; i++) {
    dp[i][0] = 1;
    long long int prevTot = dp[i - 1][0];
    for (int j = 1; j < N; j++) {
      prevTot = (prevTot + dp[i - 1][j]) % m;
      dp[i][j] = prevTot;
    }
    for (int j = N - 2; j >= 0; j--)
      if (v[j].first == v[j + 1].first) dp[i][j] = dp[i][j + 1];
  }
  long long int total = 0;
  long long int s1 = L / ((long long)N), s2 = L / ((long long)N);
  if (L % N != 0) s2 += 1LL;
  long long int c1 = s1 % m, c2 = s2 % m;
  long long int maxAccept = L % ((long long)N) - 1LL;
  for (int i = 1; i <= K && i <= s2; i++) {
    for (int j = 0; j < N; j++) {
      if (v[j].second <= maxAccept)
        total = (total + (c2 * dp[i][j]) % m) % m;
      else
        total = (total + (c1 * dp[i][j]) % m) % m;
    }
    c1 -= 1;
    c2 -= 1;
  }
  cout << (m + total) % m << endl;
  return 0;
}
