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;
struct pos {
  long long val, id;
  bool operator<(pos b) const { return val < b.val; }
};
long long n, k, l, i, j, p, nr;
long long ans, nans;
pos a[1000005];
long long dp0[1000005];
long long dp1[1000005];
int main() {
  cin >> n >> l >> k;
  if (n > l) {
    cout << l;
    return 0;
  }
  nr = l / n;
  for (i = 1; i <= n; i++) {
    cin >> a[i].val;
    a[i].id = i;
  }
  sort(a + 1, a + n + 1);
  for (i = 1; i <= n; i++) {
    dp1[i] = 1;
    if (a[i].id <= l % n) {
      ans += nr + 1;
      ans %= 1000000007;
    } else {
      ans += nr;
      ans %= 1000000007;
    }
  }
  for (j = 2; j <= k; j++) {
    if (j > nr) break;
    for (i = 1; i <= n; i++) dp0[i] = dp1[i];
    p = 0;
    nans = 0;
    for (i = 1; i <= n; i++) {
      while (p < n && a[p + 1].val <= a[i].val) {
        p++;
        nans += dp0[p];
        nans %= 1000000007;
      }
      dp1[i] = nans;
      if (a[i].id <= l % n) {
        ans += (nans * ((nr + 2 - j) % 1000000007)) % 1000000007;
        ans %= 1000000007;
      } else {
        ans += (nans * ((nr + 1 - j) % 1000000007)) % 1000000007;
        ans %= 1000000007;
      }
    }
  }
  if (nr < k) {
    for (i = 1; i <= n; i++) dp0[i] = dp1[i];
    p = 0;
    nans = 0;
    for (i = 1; i <= n; i++) {
      if (a[i].id > l % n) continue;
      while (p < n && a[p + 1].val <= a[i].val) {
        p++;
        nans += dp0[p];
        nans %= 1000000007;
      }
      ans += nans;
      ans %= 1000000007;
    }
  }
  cout << ans;
  return 0;
}
