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 long long md = 1e9 + 7;
int main() {
  int n, k, a;
  long long l;
  scanf("%d%lld%d", &n, &l, &k);
  map<int, int> kr;
  map<int, int> late;
  int gr = l % n == 0 ? n : l % n;
  for (int i = (0); i < int(n); ++i) {
    scanf("%d", &a);
    ++kr[a];
    if (i >= gr) ++late[a];
  }
  vector<int> flat, lfl;
  for (auto par : kr) {
    flat.push_back(par.second);
    lfl.push_back(late[par.first]);
  }
  int f = flat.size();
  vector<vector<long long> > dp(k, vector<long long>(f));
  for (int i = (0); i < int(f); ++i) dp[0][i] = flat[i];
  long long bnum = (1 + (l - 1) / n);
  long long ans = l % md;
  for (int i = (1); i < int(min((long long)k, bnum)); ++i) {
    long long bef = 0;
    for (int j = (0); j < int(f); ++j) {
      bef = (bef + dp[i - 1][j]) % md;
      dp[i][j] = bef * flat[j] % md;
      ans = (ans + (bnum % md - i) * dp[i][j] - bef * lfl[j]) % md;
    }
  }
  printf("%lld\n", (ans + md) % md);
}
