#include <bits/stdc++.h>
using namespace std;
template <typename T>
void arrout(T arr[], int len) {
  for (int i = 0; i <= len; ++i) cout << arr[i] << " ";
  cout << endl;
}
int OK = 1;
const int INF = 0x3f3f3f3f;
const int maxn = 1001000;
long long n, k;
long long l;
int arr[maxn], tarr[maxn];
long long mod = 1e9 + 7;
int p[maxn];
long long sum[maxn], dp[maxn];
int main() {
  scanf("%I64d%I64d%I64d", &n, &l, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", arr + i), tarr[i] = arr[i];
  sort(arr + 1, arr + 1 + n);
  for (int i = 1; i <= n; ++i)
    p[i] = lower_bound(arr + 1, arr + 1 + n, tarr[i]) - arr;
  for (int i = 1; i <= n; ++i) sum[i] = 1;
  long long ans = 0;
  long long t = l / n, tt = l % n;
  long long ttt = min(t, k);
  t++;
  ;
  ;
  if (tt) ttt++;
  ttt = min(ttt, k);
  for (int i = 1; i <= ttt; ++i) {
    for (int j = 1; j <= n; ++j) {
      dp[j] = sum[j];
      sum[j] = sum[j - 1] + dp[j];
      sum[j] %= mod;
    }
    for (int j = 1; j <= tt; ++j) {
      ans += (t - i + 1) % mod * dp[p[j]];
      ans %= mod;
    }
    for (int j = tt + 1; j <= n; ++j) {
      ans += (t - i) % mod * dp[p[j]];
      ans %= mod;
    }
    for (int j = n - 1; j >= 1; --j)
      if (arr[j] == arr[j + 1]) sum[j] = sum[j + 1];
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Generate a cpp solution to the following problem:

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>.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void arrout(T arr[], int len) {
  for (int i = 0; i <= len; ++i) cout << arr[i] << " ";
  cout << endl;
}
int OK = 1;
const int INF = 0x3f3f3f3f;
const int maxn = 1001000;
long long n, k;
long long l;
int arr[maxn], tarr[maxn];
long long mod = 1e9 + 7;
int p[maxn];
long long sum[maxn], dp[maxn];
int main() {
  scanf("%I64d%I64d%I64d", &n, &l, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", arr + i), tarr[i] = arr[i];
  sort(arr + 1, arr + 1 + n);
  for (int i = 1; i <= n; ++i)
    p[i] = lower_bound(arr + 1, arr + 1 + n, tarr[i]) - arr;
  for (int i = 1; i <= n; ++i) sum[i] = 1;
  long long ans = 0;
  long long t = l / n, tt = l % n;
  long long ttt = min(t, k);
  t++;
  ;
  ;
  if (tt) ttt++;
  ttt = min(ttt, k);
  for (int i = 1; i <= ttt; ++i) {
    for (int j = 1; j <= n; ++j) {
      dp[j] = sum[j];
      sum[j] = sum[j - 1] + dp[j];
      sum[j] %= mod;
    }
    for (int j = 1; j <= tt; ++j) {
      ans += (t - i + 1) % mod * dp[p[j]];
      ans %= mod;
    }
    for (int j = tt + 1; j <= n; ++j) {
      ans += (t - i) % mod * dp[p[j]];
      ans %= mod;
    }
    for (int j = n - 1; j >= 1; --j)
      if (arr[j] == arr[j + 1]) sum[j] = sum[j + 1];
  }
  cout << ans << endl;
  return 0;
}

```