#include <bits/stdc++.h>
using namespace std;
const int BASE = 1000LL * 1000 * 1000 + 7;
const int INF = 1000LL * 1000 * 1000 + 10;
int n, k;
long long l;
int a[1234567], sa[1234567];
map<int, long long> ma[2];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> l >> k;
  for (int i = (0), _b = (n - 1); i <= _b; i++) {
    cin >> a[i];
    sa[i] = a[i];
  }
  sort(sa, sa + n);
  sa[n] = INF;
  int now = 0, prev = 1;
  for (int i = (0), _b = (n - 1); i <= _b; i++) {
    ma[now][sa[i]]++;
    if (sa[i] < sa[i + 1]) ma[now][sa[i + 1]] = ma[now][sa[i]];
  }
  if (l <= n) {
    cout << l;
    return 0;
  }
  long long res = l % BASE;
  long long hs = ((l - 1) / n + 1) % BASE;
  for (int j = (2), _b = (min((long long)k, (l - 1) / n + 1)); j <= _b; j++) {
    prev = now;
    now = 1 - now;
    for (int i = (0), _b = (n - 1); i <= _b; i++) ma[now][a[i]] = 0;
    for (int i = (0), _b = (n - 1); i <= _b; i++) {
      (res += ma[prev][a[i]] * (hs - j) +
              ma[prev][a[i]] * (i < (l % n == 0 ? n : l % n))) %= BASE;
      (ma[now][a[i]] += ma[prev][a[i]]) %= BASE;
    }
    for (int i = (0), _b = (n - 1); i <= _b; i++)
      if (sa[i] < sa[i + 1]) (ma[now][sa[i + 1]] += ma[now][sa[i]]) %= BASE;
  }
  if (res < 0) res += BASE;
  cout << res;
}

### Prompt

In cpp, your task is to solve 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;
const int BASE = 1000LL * 1000 * 1000 + 7;
const int INF = 1000LL * 1000 * 1000 + 10;
int n, k;
long long l;
int a[1234567], sa[1234567];
map<int, long long> ma[2];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> l >> k;
  for (int i = (0), _b = (n - 1); i <= _b; i++) {
    cin >> a[i];
    sa[i] = a[i];
  }
  sort(sa, sa + n);
  sa[n] = INF;
  int now = 0, prev = 1;
  for (int i = (0), _b = (n - 1); i <= _b; i++) {
    ma[now][sa[i]]++;
    if (sa[i] < sa[i + 1]) ma[now][sa[i + 1]] = ma[now][sa[i]];
  }
  if (l <= n) {
    cout << l;
    return 0;
  }
  long long res = l % BASE;
  long long hs = ((l - 1) / n + 1) % BASE;
  for (int j = (2), _b = (min((long long)k, (l - 1) / n + 1)); j <= _b; j++) {
    prev = now;
    now = 1 - now;
    for (int i = (0), _b = (n - 1); i <= _b; i++) ma[now][a[i]] = 0;
    for (int i = (0), _b = (n - 1); i <= _b; i++) {
      (res += ma[prev][a[i]] * (hs - j) +
              ma[prev][a[i]] * (i < (l % n == 0 ? n : l % n))) %= BASE;
      (ma[now][a[i]] += ma[prev][a[i]]) %= BASE;
    }
    for (int i = (0), _b = (n - 1); i <= _b; i++)
      if (sa[i] < sa[i + 1]) (ma[now][sa[i + 1]] += ma[now][sa[i]]) %= BASE;
  }
  if (res < 0) res += BASE;
  cout << res;
}

```