#include <bits/stdc++.h>
using namespace std;
int d[1000000];
map<int, int> c, cnt;
int n, K, r;
long long l;
int main() {
  int i, j, k, M;
  map<int, int>::iterator it;
  scanf("%d%I64d%d", &n, &l, &K);
  M = 0;
  for (i = 0; i < n; i++) {
    scanf("%d", &d[i]);
    if (M < d[i]) M = d[i];
  }
  for (i = 0; i < n; i++) c[d[i]]++;
  cnt = c;
  i = 0;
  for (it = c.begin(); it != c.end(); it++) {
    it->second += i;
    if (it->second >= 1000000007) it->second -= 1000000007;
    i = it->second;
  }
  r = l % n;
  for (k = 1; k <= K && k <= l / n; k++) {
    r += c[M] * ((l / n - k + 1) % 1000000007) % 1000000007;
    if (r >= 1000000007) r -= 1000000007;
    if (k + 1 <= K) {
      for (i = l % n - 1; i >= 0; i--) {
        r += c[d[i]];
        if (r >= 1000000007) r -= 1000000007;
      }
    }
    int pp = 0;
    for (auto i : cnt) {
      pp = (pp + (long long)c[i.first] * i.second) % 1000000007;
      c[i.first] = pp;
    }
  }
  printf("%d", r);
  return 0;
}

### Prompt

Please create a solution in CPP 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;
int d[1000000];
map<int, int> c, cnt;
int n, K, r;
long long l;
int main() {
  int i, j, k, M;
  map<int, int>::iterator it;
  scanf("%d%I64d%d", &n, &l, &K);
  M = 0;
  for (i = 0; i < n; i++) {
    scanf("%d", &d[i]);
    if (M < d[i]) M = d[i];
  }
  for (i = 0; i < n; i++) c[d[i]]++;
  cnt = c;
  i = 0;
  for (it = c.begin(); it != c.end(); it++) {
    it->second += i;
    if (it->second >= 1000000007) it->second -= 1000000007;
    i = it->second;
  }
  r = l % n;
  for (k = 1; k <= K && k <= l / n; k++) {
    r += c[M] * ((l / n - k + 1) % 1000000007) % 1000000007;
    if (r >= 1000000007) r -= 1000000007;
    if (k + 1 <= K) {
      for (i = l % n - 1; i >= 0; i--) {
        r += c[d[i]];
        if (r >= 1000000007) r -= 1000000007;
      }
    }
    int pp = 0;
    for (auto i : cnt) {
      pp = (pp + (long long)c[i.first] * i.second) % 1000000007;
      c[i.first] = pp;
    }
  }
  printf("%d", r);
  return 0;
}

```