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 maxn = 1e6 + 10;
long long dp[maxn], ps[maxn];
long long mod = 1e9 + 7;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, l, k;
  cin >> n >> l >> k;
  long long m = l % n > 0 ? l % n : n, c = (l + n - 1) / n;
  vector<pair<long long, long long> > a;
  a.push_back({0, 0});
  for (int i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    a.push_back({x, i});
  }
  sort(a.begin(), a.end());
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    dp[i] = (long long)1;
    ps[i] = (ps[i - 1] + dp[i]) % mod;
    if (a[i].second <= m) {
      ans += 1LL * dp[i] * (c % mod) % mod;
    } else {
      ans += 1LL * dp[i] * ((c - (long long)1) % mod) % mod;
    }
    ans %= mod;
  }
  for (long long j = 2; j <= min(c, k); j++) {
    for (int i = 1; i <= n; i++) {
      int s = i;
      long long x = a[i].first;
      while (a[s].first == x) {
        s++;
      }
      s--;
      for (int t = i; t <= s; t++) {
        dp[t] = ps[s] % mod;
        ps[t] = (ps[t - 1] + dp[t]) % mod;
        if (a[t].second <= m) {
          ans =
              (ans + 1LL * dp[t] % mod * ((c - j + (long long)1) % mod) % mod) %
              mod;
        } else {
          ans = (ans + 1LL * dp[t] % mod * ((c - j) % mod) % mod) % mod;
        }
        ans %= mod;
      }
      i = s;
    }
  }
  cout << ans;
  return 0;
}
