#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 4, MOD = 1e9 + 7;
int arr[N], c[N], n, k, ran[N], d[N], pref[N], A[N], sum[N], ans, i, h;
long long l, m;
int main() {
  ios::sync_with_stdio();
  cin.tie(0);
  cout.tie(0);
  cin >> n >> l >> k;
  int dp[n + 5][k + 5];
  for (i = 0; i < n; i++) cin >> arr[i], c[i] = arr[i];
  sort(c, c + n);
  for (i = 0; i < n; i++) d[i] = ran[i] = lower_bound(c, c + n, arr[i]) - c;
  sort(d, d + n);
  A[n - 1] = n - 1;
  for (i = n - 2; i >= 0; i--) A[i] = (d[i] == d[i + 1] ? A[i + 1] : i);
  m = (l) / n;
  for (i = 0; i < n; i++) dp[i][1] = 1, pref[i] = i + 1;
  sum[1] = n;
  for (h = 2; h <= k; h++) {
    for (i = 0; i < n; i++) {
      dp[i][h] = pref[A[i]];
      sum[h] = (sum[h] + dp[i][h]) % MOD;
    }
    for (i = 0; i < n; i++) {
      pref[i] = dp[i][h];
      if (i) pref[i] = (pref[i] + pref[i - 1]) % MOD;
    }
  }
  if ((long long)k > m) h = (int)m;
  for (i = 1; i <= h; i++)
    ans = (1ll * ans + 1ll * sum[i] % MOD * (m % MOD - i + 1) % MOD) % MOD;
  l -= m * n;
  ans = (ans + l % MOD) % MOD;
  for (h = 2; h <= k; h++) {
    if (m + 1 >= h) {
      for (i = 0; i < l; i++) ans = (ans + dp[ran[i]][h]) % MOD;
    }
  }
  cout << (ans % MOD + MOD) % MOD;
}

### Prompt

Create a solution in Cpp for 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;
const int N = 1e6 + 4, MOD = 1e9 + 7;
int arr[N], c[N], n, k, ran[N], d[N], pref[N], A[N], sum[N], ans, i, h;
long long l, m;
int main() {
  ios::sync_with_stdio();
  cin.tie(0);
  cout.tie(0);
  cin >> n >> l >> k;
  int dp[n + 5][k + 5];
  for (i = 0; i < n; i++) cin >> arr[i], c[i] = arr[i];
  sort(c, c + n);
  for (i = 0; i < n; i++) d[i] = ran[i] = lower_bound(c, c + n, arr[i]) - c;
  sort(d, d + n);
  A[n - 1] = n - 1;
  for (i = n - 2; i >= 0; i--) A[i] = (d[i] == d[i + 1] ? A[i + 1] : i);
  m = (l) / n;
  for (i = 0; i < n; i++) dp[i][1] = 1, pref[i] = i + 1;
  sum[1] = n;
  for (h = 2; h <= k; h++) {
    for (i = 0; i < n; i++) {
      dp[i][h] = pref[A[i]];
      sum[h] = (sum[h] + dp[i][h]) % MOD;
    }
    for (i = 0; i < n; i++) {
      pref[i] = dp[i][h];
      if (i) pref[i] = (pref[i] + pref[i - 1]) % MOD;
    }
  }
  if ((long long)k > m) h = (int)m;
  for (i = 1; i <= h; i++)
    ans = (1ll * ans + 1ll * sum[i] % MOD * (m % MOD - i + 1) % MOD) % MOD;
  l -= m * n;
  ans = (ans + l % MOD) % MOD;
  for (h = 2; h <= k; h++) {
    if (m + 1 >= h) {
      for (i = 0; i < l; i++) ans = (ans + dp[ran[i]][h]) % MOD;
    }
  }
  cout << (ans % MOD + MOD) % MOD;
}

```