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, a[1000005], b[1000005], bsize, pos[1000005];
long long k, l, f[1000005], s[1000005], g[1000005];
int np(int x) {
  int l = 1, r = bsize, kq = -1;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (b[mid] == x) {
      kq = mid;
      l = mid + 1;
    } else if (b[mid] > x)
      r = mid - 1;
    else
      l = mid + 1;
  }
  return kq;
}
int main() {
  cin >> n >> l >> k;
  k = min(k, l / (long long)n + (long long)(l % n != 0));
  bsize = 0;
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), b[++bsize] = a[i];
  sort(b + 1, b + bsize + 1);
  for (int i = 1; i <= n; ++i) pos[i] = np(a[i]);
  long long res = 0;
  for (int i = 1; i <= k; ++i) {
    if (i == 1) {
      for (int j = 1; j <= n; ++j)
        f[j] = 1LL, g[pos[j]] = (g[pos[j]] + 1LL) % 1000000007;
      for (int j = 1; j <= n; ++j) s[j] = (s[j - 1] + g[j]) % 1000000007;
    } else {
      for (int j = 1; j <= n; ++j) g[j] = 0LL;
      for (int j = 1; j <= n; ++j)
        f[j] = s[pos[j]], g[pos[j]] = (g[pos[j]] + f[j]) % 1000000007;
      for (int j = 1; j <= n; ++j) s[j] = (s[j - 1] + g[j]) % 1000000007;
    }
    if (l % n == 0) {
      res =
          (res +
           ((s[n] * (((l / (long long)n) - (long long)i + 1LL) % 1000000007)) %
            1000000007)) %
          1000000007;
    } else {
      if ((l / (long long)n) >= (long long)i)
        res =
            (res +
             (s[n] * (((l / (long long)n) - (long long)i + 1LL) % 1000000007)) %
                 1000000007) %
            1000000007;
      for (int j = 1; j <= l % n; ++j) res = (res + f[j]) % 1000000007;
    }
  }
  cout << res;
}
