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 N = 1e6 + 7;
const int M = 1e6 + 7;
const int lim = 2e6;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
int n, m, k, a[N], dp[N];
long long l;
pair<int, int> c[N];
int main() {
  scanf("%d%lld%d", &n, &l, &k);
  m = n * k;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < m; i++) {
    c[i].first = a[i % n];
    c[i].second = i;
  }
  sort(c, c + m);
  long long tot = l / n, ans = l % mod;
  int yu = l % n;
  for (int i = 0; i < m; i++) {
    int id = c[i].second;
    if (id / n == 0)
      dp[id / n + 1]++;
    else
      dp[id / n + 1] = (dp[id / n + 1] + dp[id / n]) % mod;
    if (id >= l) continue;
    if (id <= l && id / n > 0) {
      long long cnt = tot - id / n;
      if (id % n <= yu - 1) cnt++;
      ans = (ans + cnt % mod * dp[id / n] % mod) % mod;
    }
  }
  printf("%lld\n", ans);
  return 0;
}
