#include <bits/stdc++.h>
using namespace std;
map<int, int> comp;
int d[1000000];
int c[2][1000000], T;
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);
  for (i = 0; i < n; i++) {
    scanf("%d", &d[i]);
    comp[d[i]] = 1;
  }
  M = 0;
  for (it = comp.begin(); it != comp.end(); it++) it->second = M++;
  for (i = 0; i < n; i++) d[i] = comp[d[i]];
  T = 0;
  for (i = 0; i < n; i++) c[T][d[i]]++;
  for (i = 1; i < M; i++) {
    c[T][i] += c[T][i - 1];
    if (c[T][i] >= 1000000007) c[T][i] -= 1000000007;
  }
  r = l % n;
  for (k = 1; k <= K && k <= l / n; k++) {
    r += c[T][M - 1] * ((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[T][d[i]];
        if (r >= 1000000007) r -= 1000000007;
      }
    }
    T = 1 - T;
    for (i = 0; i < M; i++) c[T][i] = 0;
    for (i = 0; i < n; i++) {
      c[T][d[i]] += c[1 - T][d[i]];
      if (c[T][d[i]] >= 1000000007) c[T][d[i]] -= 1000000007;
    }
    for (i = 1; i < M; i++) {
      c[T][i] += c[T][i - 1];
      if (c[T][i] >= 1000000007) c[T][i] -= 1000000007;
    }
  }
  printf("%d", r);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
map<int, int> comp;
int d[1000000];
int c[2][1000000], T;
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);
  for (i = 0; i < n; i++) {
    scanf("%d", &d[i]);
    comp[d[i]] = 1;
  }
  M = 0;
  for (it = comp.begin(); it != comp.end(); it++) it->second = M++;
  for (i = 0; i < n; i++) d[i] = comp[d[i]];
  T = 0;
  for (i = 0; i < n; i++) c[T][d[i]]++;
  for (i = 1; i < M; i++) {
    c[T][i] += c[T][i - 1];
    if (c[T][i] >= 1000000007) c[T][i] -= 1000000007;
  }
  r = l % n;
  for (k = 1; k <= K && k <= l / n; k++) {
    r += c[T][M - 1] * ((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[T][d[i]];
        if (r >= 1000000007) r -= 1000000007;
      }
    }
    T = 1 - T;
    for (i = 0; i < M; i++) c[T][i] = 0;
    for (i = 0; i < n; i++) {
      c[T][d[i]] += c[1 - T][d[i]];
      if (c[T][d[i]] >= 1000000007) c[T][d[i]] -= 1000000007;
    }
    for (i = 1; i < M; i++) {
      c[T][i] += c[T][i - 1];
      if (c[T][i] >= 1000000007) c[T][i] -= 1000000007;
    }
  }
  printf("%d", r);
  return 0;
}

```