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>
const int BASE = 1000000007;
using namespace std;
int a[1000001], A[1000001], uper[1000001];
map<int, int> Uper;
long long l, n, k, m, d[1000001];
int main() {
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    A[i] = a[i];
  }
  sort(a, a + n);
  a[n] = a[n - 1] + 1;
  for (int i = 0, j = 0; i < n; i++) {
    while (a[j] <= a[i]) j++;
    uper[i] = Uper[a[i]] = j - 1;
  }
  for (int i = 0; i < n; i++) d[i] = i + 1;
  for (long long i = n; i < min(n * k, l); i++) {
    d[i] = (d[i - (i % n > 0)] + d[i - n + uper[i % n] - i % n]) % BASE;
  }
  m = l % n;
  for (long long i = 0; i < min(k, l / n); i++) {
    m = (m + ((((l / n) - i) % BASE) * d[i * n + n - 1]) % BASE) % BASE;
    if (i == k - 1) break;
    for (long long j = 0; j < l % n; j++) {
      m = (m + d[i * n + Uper[A[j]]]) % BASE;
    }
  }
  cout << m;
  return 0;
}
