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 M = 1e6 + 5, P = 1e9 + 7;
long long n, k, m, res, l, T, ans, mi;
int sum[2][M], A[M], B[M], tmp[M];
int main() {
  cin >> n >> l >> k;
  for (int i = 1; i <= n; i++) {
    scanf("%d", A + i);
    B[i] = A[i];
  }
  sort(B + 1, B + 1 + n);
  m = unique(B + 1, B + 1 + n) - B - 1;
  for (int i = 1; i <= n; i++) A[i] = lower_bound(B + 1, B + 1 + m, A[i]) - B;
  T = l / n;
  res = l % n;
  for (int i = 1; i <= n; i++) sum[1][A[i]]++;
  ans = l % P;
  for (int i = 1; i <= m; i++) sum[1][i] += sum[1][i - 1], tmp[i] = sum[1][i];
  for (int lop = 2; lop <= min(1ll * k, T); lop++) {
    int c = lop & 1, len = (T - lop + 1) % P;
    for (int i = 1; i <= m; i++) sum[c][i] = 0;
    for (int i = 1; i <= n; i++)
      sum[c][A[i]] = (1ll * sum[c][A[i]] + 1ll * sum[!c][A[i]]) % P;
    for (int i = 1; i <= m; i++) {
      sum[c][i] = (1ll * sum[c][i] + 1ll * sum[c][i - 1]) % P;
      if (lop < k) tmp[i] = (1ll * tmp[i] + 1ll * sum[c][i]) % P;
    }
    ans = (1ll * ans + 1ll * len * sum[c][m]) % P;
  }
  if (k >= 2 && T >= 1)
    for (int i = 1; i <= res; i++) ans = (1ll * ans + 1ll * tmp[A[i]]) % P;
  printf("%d", ans);
}
