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 = 1e6 + 15, mod = 1e9 + 7;
long long dp[N][2];
int a[N], temp[N];
int find(int x, int n) {
  int l = 0, r = n - 1, mid;
  while (l < r) {
    mid = (l + r + 1) / 2;
    if (temp[mid] <= x)
      l = mid;
    else
      r = mid - 1;
  }
  return l;
}
int cc[N];
int main() {
  long long n, l, k;
  scanf("%lld%lld%lld", &n, &l, &k);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]), temp[i] = a[i];
  sort(temp, temp + n);
  int maxi = 0;
  for (int i = 0; i < n; ++i) {
    a[i] = find(a[i], n);
    ++cc[a[i]];
    maxi = max(maxi, a[i]);
  }
  long long seg = (l - 1) / n + 1;
  long long m = min(k, seg - 1);
  int bit = 0, nbit;
  long long ret = 0;
  for (int i = 0; i < m; ++i) {
    if (i == 0) {
      for (int j = 0; j <= maxi; ++j) dp[j][bit] = cc[j];
    } else {
      nbit = bit ^ 1;
      long long sum = 0;
      for (int j = 0; j <= maxi; ++j) {
        sum = (sum + dp[j][bit]) % mod;
        dp[j][nbit] = (sum * cc[j]) % mod;
      }
      bit = nbit;
    }
    for (int j = 0; j <= maxi; ++j) {
      long long temp = (seg - 1 - i) % mod;
      ret = (ret + temp * dp[j][bit]) % mod;
    }
  }
  bit = 0;
  memset(dp, 0, sizeof(dp));
  for (long long st = (seg - 1) * n; st < l; ++st) {
    int j = st % n;
    ++dp[a[j]][bit];
    ret = (ret + 1) % mod;
  }
  k = min(k, seg);
  for (int i = 1; i < k; ++i) {
    nbit = bit ^ 1;
    long long sum = 0;
    for (int j = maxi; j >= 0; --j) {
      sum = (sum + dp[j][bit]) % mod;
      dp[j][nbit] = (sum * cc[j]) % mod;
    }
    bit = nbit;
    for (int j = 0; j <= maxi; ++j) {
      ret = (ret + dp[j][bit]) % mod;
    }
  }
  ret = (ret + mod) % mod;
  cout << ret << endl;
  return 0;
}
