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 maxn = 2000000;
const int mod = 1e9 + 7;
int n, k;
int get(int i, int j) { return (i - 1) * n + j; }
long long l;
int b[maxn], f[maxn], sum[maxn], a[maxn];
int ans = 0;
pair<int, int> c[maxn];
int main() {
  scanf("%d %I64d %d", &n, &l, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), c[i] = make_pair(a[i], i);
  sort(c + 1, c + 1 + n);
  for (int i = 1; i <= n; i++) b[i] = c[i].first;
  int cur = 0;
  if (1LL * n >= l)
    printf("%I64d\n", l);
  else {
    for (int i = 1; i <= n; i++) f[i] = 1, sum[i] = i;
    for (int i = 2; i <= k; i++) {
      for (int j = 1; j <= n; j++) {
        int tmp = (int)(upper_bound(b + 1, b + 1 + n, a[j]) - b - 1);
        if (tmp) f[get(i, j)] = sum[get(i - 1, tmp)];
      }
      for (int j = 1; j <= n; j++) {
        if (j == 1)
          sum[get(i, 1)] = f[get(i, c[j].second)];
        else
          sum[get(i, j)] = sum[get(i, j - 1)] + f[get(i, c[j].second)];
        if (sum[get(i, j)] >= mod) sum[get(i, j)] -= mod;
      }
    }
    long long size = l / n;
    int res = (int)(l % n);
    for (int i = 1; i <= min(1LL * k, size + 1); i++) {
      ans += 1LL * sum[get(i, n)] * ((size - i + 1) % mod) % mod;
      if (ans >= mod) ans -= mod;
      for (int j = 1; j <= res; j++) {
        int tmp = (int)(upper_bound(b + 1, b + 1 + n, a[j]) - b - 1);
        if (tmp) {
          ans += sum[get(i - 1, tmp)];
          if (ans >= mod) ans -= mod;
        }
      }
    }
    ans += res;
    if (ans >= mod) ans -= mod;
    printf("%d\n", ans);
  }
  return 0;
}
