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;
template <class T>
bool uax(T& a, const T& b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool uin(T& a, const T& b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
long long mulmod(long long a, long long b, long long m) {
  if (!a || !b) return 0;
  if (a & 1) return (b + mulmod(a - 1, b, m)) % m;
  return (mulmod(a / 2, b, m) * 2) % m;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, k, l;
  cin >> n >> l >> k;
  long long MOD = 1e9 + 7;
  vector<int> a(n);
  vector<int> p(n);
  for (int i = 0; i < (int)(n); ++i) {
    cin >> a[i];
    p[i] = i;
  }
  sort((p).begin(), (p).end(),
       [&](const int& lhs, const int& rhs) { return a[lhs] < a[rhs]; });
  vector<vector<long long> > dp(n, vector<long long>(k + 1, 0));
  for (int i = 0; i < (int)(n); ++i) {
    dp[i][1] = 1;
  }
  for (int j = (int)(2); j <= (int)(k); ++j) {
    long long sum = 0;
    int ptr = 0;
    for (int i = 0; i < (int)(n); ++i) {
      while (ptr < n && a[p[ptr]] <= a[p[i]]) {
        sum = (sum + dp[p[ptr]][j - 1]) % MOD;
        ptr++;
      }
      dp[p[i]][j] = sum;
    }
  }
  long long c = (long long)ceil((long double)l / n);
  long long x = (long long)(l - 1) % (long long)n;
  cerr << "c: " << c << endl;
  cerr << "x: " << x << endl;
  long long ans1 = 0;
  for (int j = (int)(1); j <= (int)(min(c, k)); ++j) {
    long long sum = 0;
    for (int i = (int)(0); i <= (int)(x); ++i) {
      sum = (sum + dp[i][j]) % MOD;
    }
    ans1 = (ans1 + mulmod(sum, c - (long long)j + 1, MOD)) % MOD;
  }
  cerr << "ans1: " << ans1 << endl;
  long long ans2 = 0;
  for (int j = (int)(1); j <= (int)(min(c - 1, k)); ++j) {
    long long sum = 0;
    for (int i = (int)(x + 1); i <= (int)(n - 1); ++i) {
      sum = (sum + dp[i][j]) % MOD;
    }
    ans2 = (ans2 + mulmod(sum, c - (long long)j, MOD)) % MOD;
  }
  cerr << "ans2: " << ans2 << endl;
  cout << (ans1 + ans2) % MOD;
  return 0;
}
