#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000010;
const int mod = 1000000007;
int a[maxn], id[maxn];
int dp[2000000];
void update(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
bool cmp(int u, int v) { return a[u] < a[v]; }
int main() {
  int n, k;
  long long l;
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) id[i] = i;
  sort(id, id + n, cmp);
  for (int i = 0; i < n; i++) dp[i * (k + 1) + 1] = 1;
  for (int t = 1; t < k; t++) {
    int pre = 0;
    for (int i = 0, j; i < n; i = j) {
      for (j = i + 1; j < n && a[id[i]] == a[id[j]]; j++)
        ;
      for (int o = i; o < j; o++) update(pre, dp[o * (k + 1) + t]);
      for (int o = i; o < j; o++) dp[o * (k + 1) + t + 1] = pre;
    }
  }
  int res = 0;
  long long seg = l / n;
  for (int t = 1; t <= k; t++) {
    int sum = 0;
    for (int i = 0; i < n; i++) update(sum, dp[i * (k + 1) + t]);
    if (seg >= t) update(res, (seg - t + 1) % mod * sum % mod);
  }
  for (int i = 0; i < n; i++)
    if (id[i] < l % n) {
      for (int t = 1; t <= k; t++)
        if (seg + 1 >= t) update(res, dp[i * (k + 1) + t]);
    }
  cout << res << endl;
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 maxn = 1000010;
const int mod = 1000000007;
int a[maxn], id[maxn];
int dp[2000000];
void update(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
bool cmp(int u, int v) { return a[u] < a[v]; }
int main() {
  int n, k;
  long long l;
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) id[i] = i;
  sort(id, id + n, cmp);
  for (int i = 0; i < n; i++) dp[i * (k + 1) + 1] = 1;
  for (int t = 1; t < k; t++) {
    int pre = 0;
    for (int i = 0, j; i < n; i = j) {
      for (j = i + 1; j < n && a[id[i]] == a[id[j]]; j++)
        ;
      for (int o = i; o < j; o++) update(pre, dp[o * (k + 1) + t]);
      for (int o = i; o < j; o++) dp[o * (k + 1) + t + 1] = pre;
    }
  }
  int res = 0;
  long long seg = l / n;
  for (int t = 1; t <= k; t++) {
    int sum = 0;
    for (int i = 0; i < n; i++) update(sum, dp[i * (k + 1) + t]);
    if (seg >= t) update(res, (seg - t + 1) % mod * sum % mod);
  }
  for (int i = 0; i < n; i++)
    if (id[i] < l % n) {
      for (int t = 1; t <= k; t++)
        if (seg + 1 >= t) update(res, dp[i * (k + 1) + t]);
    }
  cout << res << endl;
  return 0;
}

```