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, a[1000005];
long long l;
vector<int> val, cou, dp[1000005];
void add(int &a, int b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
}
void setup() {
  cin >> n >> l >> k;
  for (int i = 0; i <= n - 1; i++) {
    cin >> a[i];
    val.push_back(a[i]);
  }
  sort(val.begin(), val.end());
  val.resize(unique(val.begin(), val.end()) - val.begin());
  cou.resize(val.size());
  for (int i = 0; i <= n - 1; i++) {
    dp[i].resize(k + 1);
    dp[i][1] = 1;
    a[i] = lower_bound(val.begin(), val.end(), a[i]) - val.begin();
  }
}
void xuly() {
  for (int j = 2; j <= k; j++) {
    for (int i = 0; i <= int(cou.size()) - 1; i++) cou[i] = 0;
    for (int i = 0; i <= n - 1; i++) add(cou[a[i]], dp[i][j - 1]);
    for (int i = 1; i <= int(cou.size()) - 1; i++) add(cou[i], cou[i - 1]);
    for (int i = 0; i <= n - 1; i++) dp[i][j] = cou[a[i]];
  }
  int ans = 0;
  for (int i = 0; i <= n - 1; i++)
    for (int j = 1; j <= k; j++) {
      long long m = l / n;
      if (i < l % n) m++;
      m = m - j + 1;
      if (m > 0) ans = (ans + (m % 1000000007) * dp[i][j]) % 1000000007;
    }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  setup();
  xuly();
  return 0;
}
