#include <bits/stdc++.h>
using namespace std;
void fs(long long &number) {
  bool negative = false;
  register int c;
  number = 0;
  c = getchar();
  if (c == '-') {
    negative = true;
    c = getchar();
  }
  for (; (c > 47 && c < 58); c = getchar()) number = number * 10 + c - 48;
  if (negative) number *= -1;
}
long long n, k, m, l, ans, fin;
int li[1000005];
struct node {
  long long val;
  int i;
};
node arr[1000005];
bool comp(node a, node b) {
  if (a.val < b.val) return true;
  return false;
}
long long dp[1000005], temp[1000005];
void add(int idx, long long ct) {
  long long fact = m - ct + 1;
  if (l % n && arr[idx].i > fin) fact--;
  fact %= 1000000007;
  ans = (ans + (dp[idx] * fact) % 1000000007) % 1000000007;
}
int main() {
  int i, j;
  fs(n);
  fs(l);
  fs(k);
  for (i = 0; i < n; i++) fs(arr[i].val), arr[i].i = i;
  m = l / n;
  if (l % n) m++, fin = (l - 1) % n;
  k = min(k, m);
  sort(arr, arr + n, comp);
  li[n - 1] = n - 1;
  for (j = n - 2; j >= 0; j--)
    li[j] = (arr[j].val == arr[j + 1].val) ? li[j + 1] : j;
  for (j = 0; j < n; j++) {
    dp[j] = 1LL;
    add(j, 1);
  }
  for (i = 2; i <= k; i++) {
    temp[0] = dp[0];
    for (j = 1; j < n; j++) temp[j] = (temp[j - 1] + dp[j]) % 1000000007;
    for (j = 0; j < n; j++) dp[j] = temp[li[j]], add(j, i);
  }
  printf("%I64d", ans);
}

### Prompt

Your challenge is to write 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;
void fs(long long &number) {
  bool negative = false;
  register int c;
  number = 0;
  c = getchar();
  if (c == '-') {
    negative = true;
    c = getchar();
  }
  for (; (c > 47 && c < 58); c = getchar()) number = number * 10 + c - 48;
  if (negative) number *= -1;
}
long long n, k, m, l, ans, fin;
int li[1000005];
struct node {
  long long val;
  int i;
};
node arr[1000005];
bool comp(node a, node b) {
  if (a.val < b.val) return true;
  return false;
}
long long dp[1000005], temp[1000005];
void add(int idx, long long ct) {
  long long fact = m - ct + 1;
  if (l % n && arr[idx].i > fin) fact--;
  fact %= 1000000007;
  ans = (ans + (dp[idx] * fact) % 1000000007) % 1000000007;
}
int main() {
  int i, j;
  fs(n);
  fs(l);
  fs(k);
  for (i = 0; i < n; i++) fs(arr[i].val), arr[i].i = i;
  m = l / n;
  if (l % n) m++, fin = (l - 1) % n;
  k = min(k, m);
  sort(arr, arr + n, comp);
  li[n - 1] = n - 1;
  for (j = n - 2; j >= 0; j--)
    li[j] = (arr[j].val == arr[j + 1].val) ? li[j + 1] : j;
  for (j = 0; j < n; j++) {
    dp[j] = 1LL;
    add(j, 1);
  }
  for (i = 2; i <= k; i++) {
    temp[0] = dp[0];
    for (j = 1; j < n; j++) temp[j] = (temp[j - 1] + dp[j]) % 1000000007;
    for (j = 0; j < n; j++) dp[j] = temp[li[j]], add(j, i);
  }
  printf("%I64d", ans);
}

```