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 long long MOD = 1e9 + 7;
long long n, l, k, cnt, ans, num, p, c1, c2;
long long a[1000005], b[1000005], dp[2][1000005], tem[1000005], c[1000006];
map<long long, int> mp;
int main() {
  scanf("%I64d%I64d%I64d", &n, &l, &k);
  long long x;
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &x);
    a[i] = b[i] = x;
  }
  sort(b, b + n);
  cnt = 0;
  for (int i = 0; i < n; i++)
    if (mp.find(b[i]) == mp.end()) mp[b[i]] = ++cnt;
  for (int i = 0; i < n; i++) b[i] = mp[a[i % n]];
  num = l / n;
  p = l % n;
  for (int i = 0; i < p; i++) c[i] = b[i];
  sort(b, b + n);
  sort(c, c + p);
  for (int i = 0; i < n; i++) dp[0][b[i]] = i + 1;
  ans = l % MOD;
  cnt = 0;
  c1 = num;
  c2 = num;
  k = min(k, num + (p == 0 ? 0 : 1));
  for (int i = 2; i <= k; i++) {
    cnt++;
    c2 = (num - i + 1) % MOD;
    c1 = (c2 + 1) % MOD;
    if (i <= num) {
      for (int j = 0; j < n; j++)
        if (j == 0)
          dp[cnt & 1][b[j]] = (dp[(cnt - 1) & 1][b[j]]) % MOD;
        else
          dp[cnt & 1][b[j]] =
              (dp[cnt & 1][b[j - 1]] + dp[(cnt - 1) & 1][b[j]]) % MOD;
      ans = (ans + dp[cnt & 1][b[n - 1]] * c2 % MOD) % MOD;
    }
    for (int j = 0; j < p; j++)
      if (j == 0)
        tem[c[j]] = (dp[(cnt - 1) & 1][c[j]]) % MOD;
      else
        tem[c[j]] = (tem[c[j - 1]] + dp[(cnt - 1) & 1][c[j]]) % MOD;
    if (p > 0) ans = (ans + tem[c[p - 1]] % MOD) % MOD;
  }
  printf("%I64d\n", ans);
}
