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 int T = 1e9 + 7;
long long f[1000100], sum[1000010];
int num[1000010], a[1000100];
long long n, L, k;
bool cmp(const int A, const int B) { return a[A] < a[B]; }
int main() {
  cin >> n >> L >> k;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) f[i] = 1;
  for (int i = 1; i <= n; i++) num[i] = i;
  k--;
  long long ans = L;
  ans %= T;
  sort(num + 1, num + 1 + n, cmp);
  for (int l = 1; l <= k; l++) {
    sum[0] = 0;
    for (int i = 1; i <= n; i++) {
      sum[i] = sum[i - 1] + f[num[i]];
      if (sum[i] >= T) sum[i] -= T;
    }
    int j = 1;
    for (int i = 1; i <= n; i++) {
      while (a[num[j]] <= a[num[i]] && j <= n) j++;
      j--;
      f[num[i]] = sum[j];
    }
    for (int i = 1; i <= n; i++) {
      long long len = L / n - l;
      if (L % n >= i) len++;
      len %= T;
      if (len <= 0) continue;
      ans += len * f[i] % T;
      ans %= T;
    }
  }
  cout << ans << endl;
  return 0;
}
