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;
template <typename T>
void out(T x) {
  cout << x << endl;
  exit(0);
}
const long long mod = 1e9 + 7;
const int maxn = 1e6 + 5;
long long n, l, k;
pair<int, int> a[maxn];
long long dp[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a, a + n);
  dp[0] = 1;
  long long blocks = l / n;
  long long last = l % n;
  long long ans = 0;
  for (int len = 1; len <= k && len <= blocks + 1; len++) {
    long long cur = 0;
    int j = 0;
    for (int i = 0; i < n; i++) {
      while (j < n && a[j].first <= a[i].first) {
        cur += dp[j];
        cur %= mod;
        j++;
      }
      dp[i] = cur;
    }
    for (int i = 0; i < n; i++) {
      ans += (1LL * (blocks - len + 1) % mod * dp[i] % mod);
      ans %= mod;
    }
    for (int i = 0; i < n; i++) {
      if (a[i].second < last) {
        ans += dp[i];
        ans %= mod;
      }
    }
  }
  ans %= mod;
  ans += mod;
  ans %= mod;
  out(ans);
  return 0;
}
