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;
template <typename A, typename B>
bool cmin(A &a, const B &b) {
  return a > b ? (a = b, true) : false;
}
template <typename A, typename B>
bool cmax(A &a, const B &b) {
  return a < b ? (a = b, true) : false;
}
const double PI = acos(-1);
const double EPS = 1e-7;
long long inf = sizeof(long long) == sizeof(long long) ? 2e18 : 1e9;
long long dx[] = {0, 1, 0, -1};
long long dy[] = {1, 0, -1, 0};
signed main() {
  long long n, l, m;
  cin >> n >> l >> m;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) scanf("%d", &a[i]);
  if (l < n) {
    a.resize(l);
    n = l;
  }
  vector<long long> b = a;
  sort(b.begin(), b.end());
  vector<long long> dp(n, 1);
  long long t = min(m, l / n);
  long long out = l % n;
  vector<long long> mem1(n), mem2(n);
  for (long long i = 0; i < n; i++)
    mem1[i] = upper_bound(b.begin(), b.end(), a[i]) - b.begin() - 1;
  for (long long i = 0; i < n; i++)
    mem2[i] = upper_bound(b.begin(), b.end(), b[i]) - b.begin() - 1;
  for (long long i = 0; i < t; i++) {
    long long ncr = l / n - i;
    ncr %= 1000000007;
    for (long long j = 0; j < n; j++)
      (out += dp[j] * ncr % 1000000007) %= 1000000007;
    vector<long long> d = dp;
    for (long long j = 0; j < n - 1; j++) (d[j + 1] += d[j]) %= 1000000007;
    for (long long j = n - 1; j >= 0; j--) {
      dp[j] = d[mem2[j]];
      if (i + 2 <= m && j < l % n) {
        (out += d[mem1[j]] % 1000000007) %= 1000000007;
      }
    }
  }
  cout << out << endl;
}
