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;
int n, k, len = 0;
long long L;
vector<int> a, c;
map<int, int> p;
vector<vector<long long> > t;
void init() {
  cin >> n >> L >> k;
  a.resize(n);
  c.resize(n);
  t.resize(k);
  for (int i = 0; i < k; ++i) t[i].assign(n, 0);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    c[i] = a[i];
  }
  sort(c.begin(), c.end());
  for (int i = 0; i < n; ++i)
    if (!p.count(c[i])) p[c[i]] = len++;
  for (int i = 0; i < n; ++i) a[i] = p[a[i]];
}
long long sum(int x, int y) {
  long long result = 0;
  for (; y >= 0; y = (y & (y + 1)) - 1)
    result = (result + t[x][y]) % 1000000007;
  return result;
}
void add(int x, int y, long long val) {
  for (; y < len; y = (y | (y + 1))) t[x][y] = (t[x][y] + val) % 1000000007;
}
void solve() {
  long long ans = L % 1000000007;
  for (int i = 0; i < n; ++i) add(0, a[i], 1);
  for (int size = 1; size < min((long long)k, (L + n - 1) / n); ++size) {
    long long res = 0;
    for (int i = 0; i < n; ++i) {
      long long t = sum(size - 1, a[i]);
      res = (res + t) % 1000000007;
      add(size, a[i], t);
    }
    ans += ((((L / n) - size) % 1000000007) * res) % 1000000007;
    ans %= 1000000007;
    res = 0;
    for (int i = 0; i < (L % n); ++i) {
      long long t = sum(size - 1, a[i]);
      res = (res + t) % 1000000007;
    }
    ans += res;
    ans %= 1000000007;
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  init();
  solve();
  return 0;
}
