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 long long mod = 1e9 + 7;
int n, k;
long long l;
int a[1000005];
int A[1000005];
long long dp[1000005];
long long cnt[1000005];
long long cnt2[100005];
int main() {
  scanf("%d%lld%d", &n, &l, &k);
  for (int i = int(0); i < int(n); i++) {
    scanf("%d", a + i);
    A[i] = a[i];
  }
  sort(A, A + n);
  int _A = unique(A, A + n) - A;
  long long ans = 0;
  long long p = l / n, q = l % n;
  for (int i = int(0); i < int(n); i++) {
    a[i] = lower_bound(A, A + _A, a[i]) - A;
    cnt[a[i]]++;
    if (i < q) cnt2[a[i]]++;
    dp[a[i]]++;
  }
  ans = l % mod;
  for (long long x = 2; x <= k && x <= p + 1; x++) {
    long long sum = 0, tmp, add = 0, oadd = 0;
    for (int i = 0; i < _A; i++) {
      sum = (sum + dp[i]) % mod;
      dp[i] = sum * cnt[i] % mod;
      add = (add + dp[i]) % mod;
      oadd = (oadd + sum * cnt2[i] % mod) % mod;
    }
    ans = (ans + oadd) % mod;
    if (x <= p) ans = (ans + (p - x + 1) % mod * add % mod) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}
