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;
map<long long, long long> s, g;
long long a[1000006], b[1000006], num[1000006], mod = 1e9 + 7, n, l, k;
unordered_map<long long, long long> d[1000006], c[1000006];
int main() {
  cin >> n >> l >> k;
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    s[a[i]]++;
  }
  long long tot = 1;
  for (auto x : s) {
    b[tot] = x.first;
    num[tot++] = x.second;
  }
  long long sz = n;
  n = tot - 1;
  for (long long j = 1; j <= n; j++) {
    d[1][j] = num[j];
  }
  c[1][1] = d[1][1];
  for (long long j = 2; j <= n; j++) {
    c[1][j] = c[1][j - 1] + d[1][j];
  }
  for (long long i = 2; i <= k; i++) {
    for (long long j = 1; j <= n; j++) {
      d[i][j] = (c[i - 1][j] % mod * num[j] % mod) % mod;
    }
    c[i][1] = d[i][1] % mod;
    for (long long j = 2; j <= n; j++) {
      c[i][j] = (c[i][j - 1] % mod + d[i][j] % mod) % mod;
    }
  }
  long long v = l / sz, p = l % sz, x = (l + sz - 1) / sz, ans = l % mod;
  for (long long i = 1; i <= p; i++) {
    long long pos = upper_bound(b + 1, b + 1 + n, a[i]) - b - 1;
    g[pos]++;
  }
  for (long long i = 2; i <= k && i <= x; i++) {
    if (i <= v) {
      long long k = (v - i + 1) % mod;
      ans = (ans % mod + c[i][n] % mod * k % mod) % mod;
    }
    for (auto j : g) {
      ans = (ans % mod + c[i - 1][j.first] % mod * j.second % mod) % mod;
    }
  }
  cout << ans % mod << endl;
}
