#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6;
const long long mod = 1e9 + 7;
int arr[MAXN + 5], temp[MAXN + 5], n;
int occ[MAXN + 5];
vector<vector<long long> > dp, acc;
int R;
void Rank() {
  for (int i = 0; i < n; i++) temp[i] = arr[i];
  sort(temp, temp + n);
  map<int, int> rnk;
  R = 1;
  for (int i = 0; i < n; i++) {
    if (rnk.count(temp[i]) == 0) rnk[temp[i]] = R++;
  }
  for (int i = 0; i < n; i++) arr[i] = rnk[arr[i]];
}
int main() {
  int k;
  long long l;
  scanf("%d%I64d%d", &n, &l, &k);
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
  Rank();
  for (int i = 0; i < n; i++) occ[arr[i]]++;
  dp.resize(R + 5);
  acc.resize(R + 5);
  for (int i = 0; i < dp.size(); i++) dp[i].resize(k + 5);
  for (int i = 0; i < acc.size(); i++) acc[i].resize(k + 5);
  for (int val = 1; val < R; val++) {
    dp[val][1] = 1;
    acc[val][1] = (dp[val][1] * occ[val] % mod + acc[val - 1][1]) % mod;
  }
  for (int len = 2; len <= k; len++) {
    for (int val = 1; val < R; val++) {
      dp[val][len] = acc[val][len - 1];
      acc[val][len] = (dp[val][len] * occ[val] % mod + acc[val - 1][len]) % mod;
    }
  }
  long long res = 0;
  for (int i = 0; i < n; i++) {
    for (int len = 1; len <= k; len++) {
      long long periods = l / n;
      int last_sz = l % n;
      if (i < last_sz) periods++;
      if (len > periods) continue;
      long long cur = (dp[arr[i]][len] * ((periods - len + 1) % mod)) % mod;
      res += cur;
      res %= mod;
    }
  }
  printf("%I64d\n", res);
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 = 1e6;
const long long mod = 1e9 + 7;
int arr[MAXN + 5], temp[MAXN + 5], n;
int occ[MAXN + 5];
vector<vector<long long> > dp, acc;
int R;
void Rank() {
  for (int i = 0; i < n; i++) temp[i] = arr[i];
  sort(temp, temp + n);
  map<int, int> rnk;
  R = 1;
  for (int i = 0; i < n; i++) {
    if (rnk.count(temp[i]) == 0) rnk[temp[i]] = R++;
  }
  for (int i = 0; i < n; i++) arr[i] = rnk[arr[i]];
}
int main() {
  int k;
  long long l;
  scanf("%d%I64d%d", &n, &l, &k);
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
  Rank();
  for (int i = 0; i < n; i++) occ[arr[i]]++;
  dp.resize(R + 5);
  acc.resize(R + 5);
  for (int i = 0; i < dp.size(); i++) dp[i].resize(k + 5);
  for (int i = 0; i < acc.size(); i++) acc[i].resize(k + 5);
  for (int val = 1; val < R; val++) {
    dp[val][1] = 1;
    acc[val][1] = (dp[val][1] * occ[val] % mod + acc[val - 1][1]) % mod;
  }
  for (int len = 2; len <= k; len++) {
    for (int val = 1; val < R; val++) {
      dp[val][len] = acc[val][len - 1];
      acc[val][len] = (dp[val][len] * occ[val] % mod + acc[val - 1][len]) % mod;
    }
  }
  long long res = 0;
  for (int i = 0; i < n; i++) {
    for (int len = 1; len <= k; len++) {
      long long periods = l / n;
      int last_sz = l % n;
      if (i < last_sz) periods++;
      if (len > periods) continue;
      long long cur = (dp[arr[i]][len] * ((periods - len + 1) % mod)) % mod;
      res += cur;
      res %= mod;
    }
  }
  printf("%I64d\n", res);
}

```