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;
long long int cnt[1000000 + 1];
int a[1000000 + 1];
map<int, int> mp;
void Solve(int testcase);
int main() {
  int tc, t;
  tc = 1;
  for (t = 1; t <= tc; t++) Solve(t);
  return 0;
}
void Solve(int testcase) {
  long long int n, k;
  long long int l, layers, rem;
  long long int mod = 1000000007;
  scanf("%lld %lld %lld", &n, &l, &k);
  layers = (l + n - 1) / n;
  rem = l % n;
  if (rem == 0) rem = n;
  k = min(k, layers);
  for (int i = 1; i <= n; i++) {
    a[i] = i;
    scanf("%d", &a[i]);
    mp[a[i]] = 1;
  }
  int cont = 0;
  for (map<int, int>::iterator it = mp.begin(); it != mp.end(); it++) {
    cont++;
    (*it).second = cont;
  }
  for (int i = 1; i <= n; i++) {
    a[i] = mp[a[i]];
    cnt[a[i]]++;
  }
  long long int **dp;
  dp = new long long int *[k + 1];
  for (int i = 0; i <= k; i++) {
    dp[i] = new long long int[n + 1];
    for (int j = 0; j <= n; j++) {
      dp[i][j] = 0;
    }
  }
  for (int j = 0; j <= n; j++) dp[0][j] = 1;
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= n; j++) {
      dp[i][j] = (dp[i - 1][j] * cnt[j] + dp[i][j - 1]) % mod;
    }
  }
  long long int ans = 0;
  for (int j = 1; j <= k; j++) {
    long long int md = (layers - j) % mod;
    ans = (ans + 1ll * dp[j][n] * md) % mod;
  }
  for (int j = 1; j <= k; j++) {
    for (int i = 1; i <= rem; i++) {
      ans = (ans + dp[j - 1][a[i]]) % mod;
    }
  }
  printf("%lld\n", ans);
}
