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 = 1e9 + 7;
const int N = 1e6 + 5;
long long dp[2][N], a[N], sum[N];
map<long long, long long> ma;
map<long long, long long>::iterator it;
int main() {
  long long n, l, k;
  scanf("%lld%lld%lld", &n, &l, &k);
  long long kuai = l / n;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    ma[a[i]] = 0;
  }
  int top = 0;
  for (it = ma.begin(); it != ma.end(); ++it) it->second = top++;
  for (int i = 0; i < n; i++) a[i] = ma[a[i]];
  long long pi = min(k, kuai);
  for (int i = 0; i < n; i++) dp[0][a[i]]++;
  for (int i = 0; i < top; i++) dp[0][i] += dp[0][i - 1];
  for (int i = 0; i < top; i++) sum[i] = dp[0][i];
  int u = 1, v = 0;
  long long ans = l % mod;
  long long res = l % n;
  for (long long s = 2; s <= pi; s++) {
    for (int i = 0; i < top; i++) dp[u][i] = 0;
    for (int i = 0; i < n; i++) dp[u][a[i]] = (dp[u][a[i]] + dp[v][a[i]]) % mod;
    for (int i = 0; i < top; i++) dp[u][i] = (dp[u][i] + dp[u][i - 1]) % mod;
    long long dd = (kuai - s + 1) % mod;
    if (s < k)
      for (int i = 0; i < top; i++) sum[i] = (sum[i] + dp[u][i]) % mod;
    ans = (ans + (dp[u][top - 1] * dd) % mod) % mod;
    u ^= 1, v ^= 1;
  }
  if (res && kuai >= 1 && k >= 2)
    for (int i = 0; i < res; i++) ans = (ans + sum[a[i]]) % mod;
  printf("%lld\n", ans);
}
