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;
pair<long long, long long> arr[4000006];
long long inp[4000006];
long long dp[4000006];
int main() {
  long long n, l, k;
  scanf("%lld", &n);
  ;
  scanf("%lld", &l);
  ;
  scanf("%lld", &k);
  ;
  long long bl = l / n;
  if (l % n) bl++;
  long long t1, t2;
  for (long long i = 1; i < n + 1; i++) {
    scanf("%lld", &arr[i].first);
    ;
    arr[i].second = i;
  }
  sort(arr + 1, arr + n + 1);
  for (long long i = 0; i < (n + 2) * (k + 2); i++) dp[i] = 0;
  for (long long i = 1; i < n + 1; i++) {
    long long j = i;
    while (j <= n && arr[j].first == arr[i].first) j++;
    for (long long o = i; o < j + 1; o++) inp[arr[o].second] = j - 1;
    i = j - 1;
  }
  for (long long j = 1; j < n + 1; j++) dp[j] = 1;
  for (long long i = 1; i < k + 3; i++) {
    dp[i * n + 1] = dp[(i - 1) * n + 1];
    for (long long j = 2; j < n + 1; j++) {
      dp[i * n + j] = dp[(i - 1) * n + j] + dp[i * n + j - 1];
      if (dp[i * n + j] >= 1000000007LL) dp[i * n + j] -= 1000000007LL;
      if (j == n || arr[j].first != arr[j + 1].first) {
        long long o = j, tmp = arr[j].first;
        while (o > 0 && arr[o].first == tmp) {
          dp[i * n + o] = dp[i * n + j];
          o--;
        }
      }
    }
  }
  long long co = 0LL;
  for (long long i = 0; i < k; i++) {
    if (bl <= i) break;
    co += (dp[(i + 2) * n] * ((bl - i) % 1000000007LL)) % 1000000007LL;
    if (co >= 1000000007LL) co -= 1000000007LL;
    if (l % n == 0) continue;
    for (long long j = n; j > (l % n); j--) {
      long long po = inp[j];
      co -= dp[i * n + po];
      if (co < 0) co += 1000000007LL;
    }
  }
  cout << co << endl;
}
