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 nums[1000006];
vector<int> dp[1000006];
vector<int> cum_S[1000006];
int N, K;
long long L;
int st[1000006], en[1000006];
const int mod = 1000000007;
int REM[1000006];
map<int, int> MP;
int main() {
  int i, j, k, l, rem;
  cin >> N >> L >> K;
  long long seg;
  seg = L / N;
  rem = (int)(L % N);
  MP.clear();
  for (i = 1; i <= N; i++) {
    scanf("%d", &nums[i]);
  }
  for (i = 1; i <= rem; i++) REM[i] = nums[i];
  sort(nums + 1, nums + N + 1);
  for (i = 1; i <= N; i++) MP[nums[i]] = i;
  if (rem != 0) sort(REM + 1, REM + rem + 1);
  for (i = 1; i <= N; i++) {
    j = i + 1;
    while (j <= N && nums[i] == nums[j]) j++;
    for (k = i; k < j; k++) {
      st[k] = i;
      en[k] = j - 1;
    }
    i = j - 1;
  }
  for (i = 0; i <= 1000000; i++) {
    dp[i].clear();
    cum_S[i].clear();
  }
  for (i = 0; i <= K; i++) {
    dp[i].resize(N + 1);
    cum_S[i].resize(N + 1);
  }
  for (i = 0; i <= K; i++) {
    for (j = 0; j <= N; j++) {
      dp[i][j] = cum_S[i][j] = 0;
    }
  }
  for (i = 1; i <= N; i++) {
    dp[1][i] = 1;
    cum_S[1][i] = i;
  }
  for (i = 2; i <= K; i++) {
    cum_S[i][0] = 0;
    for (j = 1; j <= N; j++) {
      dp[i][j] = cum_S[i - 1][en[j]];
      cum_S[i][j] = (cum_S[i][j - 1] + dp[i][j]) % mod;
    }
  }
  long long tmp;
  long long ans = 0;
  for (i = 1; i <= K; i++) {
    if (seg - i + 1 <= 0) break;
    tmp = ((long long)cum_S[i][N] * (long long)((seg - i + 1) % mod)) % mod;
    ans = (ans + tmp) % mod;
  }
  ans = (ans + rem) % mod;
  int limit = K;
  if (rem != 0) seg++;
  if (seg < (long long)K) limit = (int)seg;
  for (i = 1; i <= rem; i++) {
    int id = MP[REM[i]];
    for (l = 2; l <= limit; l++) {
      ans = (ans + cum_S[l - 1][en[id]]) % mod;
    }
  }
  cout << ans << endl;
  return 0;
}
