#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-10;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
long long dp[2][1000010];
long long a[1000010], b[1000010];
int n, k;
long long l;
int main() {
  cin >> n >> l >> k;
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d", &a[i]);
    b[i] = a[i];
  }
  sort(b + 1, b + 1 + n);
  int cnt = unique(b + 1, b + 1 + n) - b;
  for (int i = 1; i <= n; ++i) {
    a[i] = lower_bound(b + 1, b + cnt, a[i]) - b;
  }
  int now = 1, la = 0;
  dp[now][0] = 1;
  long long ans = 0;
  long long seg = l / n;
  for (int i = 1; i <= k && i <= seg + 1; ++i) {
    swap(la, now);
    for (int j = 0; j <= cnt; ++j) dp[now][j] = 0;
    for (int j = 1; j <= cnt; ++j)
      dp[la][j] = (dp[la][j - 1] + dp[la][j]) % MOD;
    for (int j = 1; j <= n; ++j)
      dp[now][a[j]] = (dp[now][a[j]] + dp[la][a[j]]) % MOD;
    for (int j = 1; j <= cnt; ++j)
      ans = (ans + (seg - i + 1) % MOD * dp[now][j] % MOD) % MOD;
    for (int j = 1; j <= l - seg * n; ++j) ans = (ans + dp[la][a[j]]) % MOD;
  }
  cout << ans << 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 double eps = 1e-10;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
long long dp[2][1000010];
long long a[1000010], b[1000010];
int n, k;
long long l;
int main() {
  cin >> n >> l >> k;
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d", &a[i]);
    b[i] = a[i];
  }
  sort(b + 1, b + 1 + n);
  int cnt = unique(b + 1, b + 1 + n) - b;
  for (int i = 1; i <= n; ++i) {
    a[i] = lower_bound(b + 1, b + cnt, a[i]) - b;
  }
  int now = 1, la = 0;
  dp[now][0] = 1;
  long long ans = 0;
  long long seg = l / n;
  for (int i = 1; i <= k && i <= seg + 1; ++i) {
    swap(la, now);
    for (int j = 0; j <= cnt; ++j) dp[now][j] = 0;
    for (int j = 1; j <= cnt; ++j)
      dp[la][j] = (dp[la][j - 1] + dp[la][j]) % MOD;
    for (int j = 1; j <= n; ++j)
      dp[now][a[j]] = (dp[now][a[j]] + dp[la][a[j]]) % MOD;
    for (int j = 1; j <= cnt; ++j)
      ans = (ans + (seg - i + 1) % MOD * dp[now][j] % MOD) % MOD;
    for (int j = 1; j <= l - seg * n; ++j) ans = (ans + dp[la][a[j]]) % MOD;
  }
  cout << ans << endl;
  return 0;
}

```