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 node {
  unsigned long long w;
  int no;
  bool operator<(const node &B) const { return w < B.w; }
};
int n, k;
unsigned long long l;
node A[1000007];
unsigned long long B[1000007] = {0};
unsigned long long dp[2][1000007] = {0};
int main() {
  scanf("%d%llu%d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%llu", &A[i].w);
    A[i].no = i;
    B[i] = A[i].w;
  }
  sort(A, A + n);
  A[n].w = 0;
  unsigned long long c = l / n;
  int k0 = min(c, (unsigned long long)k);
  int m = l % n;
  unsigned long long ans = 0;
  if (!c) goto mark;
  for (int i = 0; i < n; i++) {
    dp[0][i] = i + 1;
    if (A[i].no < m)
      ans = (ans + c + 1) % 1000000007ULL;
    else
      ans = (ans + c) % 1000000007ULL;
  }
  for (unsigned long long i = 1; i < k0; i++) {
    for (int j = n - 1; j >= 0; j--) {
      if (A[j].w == A[j + 1].w)
        dp[1][j] = dp[1][j + 1];
      else
        dp[1][j] = dp[0][j];
      if (A[j].no < m)
        ans = (ans + ((c - i + 1) % 1000000007ULL) * dp[1][j]) % 1000000007ULL;
      else
        ans = (ans + ((c - i) % 1000000007ULL) * dp[1][j]) % 1000000007ULL;
    }
    dp[0][0] = dp[1][0];
    for (int j = 1; j < n; j++)
      dp[0][j] = (dp[0][j - 1] + dp[1][j]) % 1000000007ULL;
  }
  if (c <= (unsigned long long)(k - 1)) {
    for (int j = n - 1; j >= 0; j--) {
      if (A[j].w == A[j + 1].w)
        dp[1][j] = dp[1][j + 1];
      else
        dp[1][j] = dp[0][j];
      if (A[j].no < m) ans += dp[1][j];
      ans %= 1000000007ULL;
    }
  }
  ans %= 1000000007ULL;
mark:
  if (!c) ans = l;
  printf("%llu\n", ans);
  return 0;
}
