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 n, k, a[1000100];
int link[1000100];
long long l;
vector<vector<long long> > v, t;
vector<pair<long long, long long> > s;
long long ans;
long long mod(long long a, long long b) {
  if (b == 0) return 0;
  if (b % 2 == 0) return (2 * mod(a, b / 2)) % 1000000007;
  return (a + mod(a, b - 1)) % 1000000007;
}
int main() {
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    int q;
    scanf("%i", &q);
    a[i] = (long long)q;
    vector<long long> temp;
    temp.assign(k + 5, 0);
    v.push_back(temp);
    t.push_back(temp);
    s.push_back(make_pair(a[i], i));
  }
  sort(s.begin(), s.end());
  link[s.size() - 1] = s.size() - 1;
  int temp = s.size() - 1;
  for (int i = s.size() - 2; i >= 0; i--) {
    if (s[i].first < s[i + 1].first) temp = i;
    link[i] = temp;
  }
  for (long long i = 0; i < n; i++) {
    v[i][1] = 1;
    t[i][1] = i + 1;
  }
  for (long long i = 2; i <= k; i++) {
    for (long long j = 0; j < n; j++) {
      v[s[j].second][i] = t[link[j]][i - 1] % 1000000007;
    }
    for (long long j = 0; j < n; j++) {
      if (j > 0)
        t[j][i] = (v[s[j].second][i] + t[j - 1][i]) % 1000000007;
      else
        t[j][i] = v[s[j].second][i] % 1000000007;
    }
  }
  for (long long i = 0; i < l % n; i++) {
    for (long long j = 1; j <= min(k, (l - 1) / n + 1); j++) {
      ans = (ans + v[i][j]) % 1000000007;
    }
  }
  long long sz = l / n;
  for (long long i = 1; i <= min(k, sz); i++) {
    long long temp = 0;
    for (long long j = 0; j < n; j++) {
      temp = (temp + v[j][i]);
    }
    temp = mod(temp, sz - i + 1) % 1000000007;
    ans = (ans + temp) % 1000000007;
  }
  ans = ans % 1000000007;
  cout << ans;
  return 0;
}
