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 long long mod = 1000000007;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
const long long N = 1000006;
long long a[N];
pair<long long, long long> e[N];
signed main() {
  ios ::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(10);
  cout << fixed;
  long long n, L, K;
  cin >> n >> L >> K;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (long long i = 0; i < n; i++) {
    e[i] = make_pair(a[i], i);
  }
  sort(e, e + n);
  long long dp[n][K + 1];
  memset(dp, 0, sizeof dp);
  for (long long i = 0; i < n; i++) {
    dp[i][1] = 1;
  }
  for (long long k = 2; k <= K; k++) {
    long long pointer = 0, sum = 0;
    for (long long i = 0; i < n; i++) {
      while (pointer < n && e[i].first >= e[pointer].first) {
        sum += dp[e[pointer++].second][k - 1];
        sum %= mod;
      }
      dp[e[i].second][k] = sum;
    }
  }
  long long ans = 0;
  long long c = (L + n - 1) / n;
  long long x = (L - 1) % n;
  for (long long i = 0; i <= x; i++) {
    for (long long j = 1; j <= min(c, K); j++) {
      ans += 1LL * dp[i][j] * ((c - j + 1) % mod);
      ans %= mod;
    }
  }
  for (long long i = x + 1; i <= n - 1; i++) {
    for (long long j = 1; j <= min(c - 1, K); j++) {
      ans += 1LL * dp[i][j] * ((c - j) % mod);
      ans %= mod;
    }
  }
  cout << (ans) << endl;
}
