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;
long long gcd(long long u, long long v) { return (v != 0) ? gcd(v, u % v) : u; }
int main() {
  long long n, k, l;
  cin >> n >> l >> k;
  vector<pair<long long, long long> > arr;
  long long i, j;
  for (i = 0; i < n; ++i) {
    long long a;
    scanf("%I64d", &a);
    arr.push_back(make_pair(a, i));
  }
  sort(arr.begin(), arr.end());
  vector<vector<long long> > dp(n + 2, vector<long long>(k + 2, 0));
  for (i = 0; i < n; ++i) dp[i][1] = 1;
  for (long long x = 2; x <= k; ++x) {
    j = 0;
    long long sum = 0;
    for (i = 0; i < n; ++i) {
      while (j < n && arr[j].first <= arr[i].first) {
        sum = (sum + dp[j][x - 1]) % 1000000007;
        j++;
      }
      dp[i][x] = sum;
    }
  }
  long long m = l / (long long)n;
  long long rem = l % (long long)n;
  long long ans = 0;
  for (long long x = 1; x <= k; ++x) {
    if (x > m + 1) break;
    long long tt = (m - x + 1) % 1000000007;
    for (i = 0; i < n; ++i) {
      long long pos = arr[i].second;
      if (x == m + 1) {
        if (pos < rem) ans = (ans + dp[i][x]) % 1000000007;
      } else {
        long long s = (dp[i][x] * tt) % 1000000007;
        ans = (ans + s) % 1000000007;
        if (pos < rem) ans = (ans + dp[i][x]) % 1000000007;
      }
    }
  }
  cout << ans;
  return 0;
}
