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;
const int N = 1000010;
const int mod = 1000000007;
map<int, int> mp;
int a[N], b[N], pp[N], p[N];
long long dp[N + N], s[N + N];
int main() {
  int n, k;
  long long l;
  scanf("%d%I64d%d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  sort(b, b + n);
  int tt = 1;
  for (int i = 0; i < n; i++) {
    if (mp[b[i]] == 0) mp[b[i]] = tt++;
  }
  for (int i = 0; i < n; i++) a[i] = mp[a[i]];
  for (int i = 0; i < n; i++) pp[a[i]]++;
  int nn = n;
  n = tt - 1;
  for (int i = 1; i <= n; i++) {
    dp[i * (k + 1) + 1] = pp[i];
    s[i * (k + 1) + 1] = s[(i - 1) * (k + 1) + 1] + pp[i];
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 2; j <= k; j++) {
      dp[i * (k + 1) + j] = (1ll * pp[i] * s[i * (k + 1) + j - 1]) % mod;
      s[i * (k + 1) + j] =
          (dp[i * (k + 1) + j] + s[(i - 1) * (k + 1) + j]) % mod;
    }
  }
  long long ans = 0;
  long long t = l / nn;
  for (int i = 1; i <= k; i++) {
    if (i > t) break;
    ans += (1ll * (t - i + 1) % mod) * s[n * (k + 1) + i] % mod;
    ans %= mod;
  }
  long long r = l % nn;
  if (r != 0) {
    ans += r;
    ans %= mod;
    for (int i = 0; i < r; i++) {
      for (int j = 1; j < k; j++) {
        if (j > t) break;
        ans += s[a[i] * (k + 1) + j];
        ans %= mod;
      }
    }
  }
  printf("%I64d\n", ans);
  return 0;
}
