#include <bits/stdc++.h>
using namespace std;
int n, K, a[1100000], x[1100000], dp[2][1100000], ans;
long long l;
const int mo = 1e9 + 7;
int compare(int k1, int k2) { return a[k1] < a[k2]; }
int calc(int K, int now) {
  int len = (l / n) % mo;
  int ans = 0;
  if (l / n + 1 >= K)
    for (int i = 1; i <= l % n; i++)
      ans = (ans + 1ll * dp[now][i] * (len - K + 2)) % mo;
  if (l / n >= K)
    for (int i = l % n + 1; i <= n; i++)
      ans = (ans + 1ll * dp[now][i] * (len - K + 1)) % mo;
  return ans;
}
int main() {
  scanf("%d%I64d%d", &n, &l, &K);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) x[i] = i;
  sort(x + 1, x + n + 1, compare);
  int now = 0;
  for (int i = 1; i <= n; i++) dp[0][i] = 1;
  ans = (ans + calc(1, 0)) % mo;
  for (int i = 2; i <= K; i++) {
    int ne = (now ^ 1);
    for (int j = 1; j <= n; j++) dp[ne][j] = 0;
    int pre = 1, tot = dp[now][x[1]];
    for (int j = 2; j <= n + 1; j++) {
      if (j > n || a[x[j]] != a[x[pre]]) {
        for (int k = pre; k < j; k++) dp[ne][x[k]] = tot;
        pre = j;
      }
      tot = (tot + dp[now][x[j]]) % mo;
    }
    now ^= 1;
    ans = (ans + calc(i, now)) % mo;
  }
  printf("%d\n", (ans + mo) % mo);
  return 0;
}

### Prompt

Generate 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;
int n, K, a[1100000], x[1100000], dp[2][1100000], ans;
long long l;
const int mo = 1e9 + 7;
int compare(int k1, int k2) { return a[k1] < a[k2]; }
int calc(int K, int now) {
  int len = (l / n) % mo;
  int ans = 0;
  if (l / n + 1 >= K)
    for (int i = 1; i <= l % n; i++)
      ans = (ans + 1ll * dp[now][i] * (len - K + 2)) % mo;
  if (l / n >= K)
    for (int i = l % n + 1; i <= n; i++)
      ans = (ans + 1ll * dp[now][i] * (len - K + 1)) % mo;
  return ans;
}
int main() {
  scanf("%d%I64d%d", &n, &l, &K);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) x[i] = i;
  sort(x + 1, x + n + 1, compare);
  int now = 0;
  for (int i = 1; i <= n; i++) dp[0][i] = 1;
  ans = (ans + calc(1, 0)) % mo;
  for (int i = 2; i <= K; i++) {
    int ne = (now ^ 1);
    for (int j = 1; j <= n; j++) dp[ne][j] = 0;
    int pre = 1, tot = dp[now][x[1]];
    for (int j = 2; j <= n + 1; j++) {
      if (j > n || a[x[j]] != a[x[pre]]) {
        for (int k = pre; k < j; k++) dp[ne][x[k]] = tot;
        pre = j;
      }
      tot = (tot + dp[now][x[j]]) % mo;
    }
    now ^= 1;
    ans = (ans + calc(i, now)) % mo;
  }
  printf("%d\n", (ans + mo) % mo);
  return 0;
}

```