#include <bits/stdc++.h>
using namespace std;
const int N = 1000006;
const int MOD = 1000000007;
long long l;
int n, k, b[N], S[N], ans;
pair<int, int> a[N];
int mul(int a, long long b) {
  if (a <= 1 || b <= 1) return (a * b) % MOD;
  a = a % MOD;
  return (mul(a * 2, b / 2) + a * (b % 2)) % MOD;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> l >> k;
  for (int i = 1; i <= n; i++) cin >> a[i].first, a[i].second = i;
  sort(a + 1, a + n + 1);
  for (int i = n; i > 0; i--)
    if (a[i].first == a[i + 1].first)
      b[i] = b[i + 1] + 1;
    else
      b[i] = 1;
  for (int i = 1; i <= n; i++) S[i] = S[i - 1] + 1;
  long long num = ((l - 1) / n) + 1;
  for (int j = 2; j <= k; j++) {
    for (int i = 1; i <= n; i++) {
      int p = (j - 1) * n + i;
      int tmp = S[p - n + b[i] - 1];
      if (i == 1)
        S[p] = tmp;
      else
        S[p] = (S[p - 1] + tmp) % MOD;
      if (a[i].second + (j - 1) * n <= l) {
        long long first = num - j + 1;
        if (1ll * (first + j - 2) * n + a[i].second > l) first--;
        ans = (ans + mul(tmp, first)) % MOD;
      }
    }
  }
  l = l % MOD;
  cout << (ans + l) % MOD;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
const int N = 1000006;
const int MOD = 1000000007;
long long l;
int n, k, b[N], S[N], ans;
pair<int, int> a[N];
int mul(int a, long long b) {
  if (a <= 1 || b <= 1) return (a * b) % MOD;
  a = a % MOD;
  return (mul(a * 2, b / 2) + a * (b % 2)) % MOD;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> l >> k;
  for (int i = 1; i <= n; i++) cin >> a[i].first, a[i].second = i;
  sort(a + 1, a + n + 1);
  for (int i = n; i > 0; i--)
    if (a[i].first == a[i + 1].first)
      b[i] = b[i + 1] + 1;
    else
      b[i] = 1;
  for (int i = 1; i <= n; i++) S[i] = S[i - 1] + 1;
  long long num = ((l - 1) / n) + 1;
  for (int j = 2; j <= k; j++) {
    for (int i = 1; i <= n; i++) {
      int p = (j - 1) * n + i;
      int tmp = S[p - n + b[i] - 1];
      if (i == 1)
        S[p] = tmp;
      else
        S[p] = (S[p - 1] + tmp) % MOD;
      if (a[i].second + (j - 1) * n <= l) {
        long long first = num - j + 1;
        if (1ll * (first + j - 2) * n + a[i].second > l) first--;
        ans = (ans + mul(tmp, first)) % MOD;
      }
    }
  }
  l = l % MOD;
  cout << (ans + l) % MOD;
  return 0;
}

```