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 mod = 1e9 + 7;
const int maxn = 1e6 + 9;
int a[maxn];
pair<int, int> pp[maxn];
int dp[maxn];
void add(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
int main() {
  int m, k;
  long long int l;
  scanf("%d %I64d %d", &m, &l, &k);
  for (int i = 0; i < m; i++) {
    scanf("%d", &a[i]);
  }
  int n = m * k;
  long long int ans = l % mod;
  for (int i = 0; i < n; i++) {
    pp[i].first = a[i % m];
    pp[i].second = i;
  }
  sort(pp, pp + n);
  long long int tot = l / m;
  int yu = l % m;
  for (int i = 0; i < n; ++i) {
    int id = pp[i].second;
    if (id / m + 1 == 1) {
      dp[id / m + 1]++;
      continue;
    }
    dp[id / m + 1] += dp[id / m + 1 - 1], dp[id / m + 1] %= mod;
    if (id >= l) continue;
    long long int cnt = tot - (id / m + 1) + 1;
    if (id % m <= yu - 1) cnt++;
    ans += ((long long int)dp[id / m] % mod) * (cnt % mod) % mod;
    ans %= mod;
  }
  printf("%I64d\n", ans);
  return 0;
}
