#include <bits/stdc++.h>
using namespace std;
int n, k;
long long l;
int a[1000000];
long long dp[2000000];
map<int, int> cnt, id;
int main() {
  scanf("%d %I64d %d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    cnt[a[i]]++;
  }
  if (k == 1) {
    printf("%I64d\n", l % 1000000007);
    return 0;
  }
  if (l <= n) {
    printf("%I64d\n", l);
    return 0;
  }
  vector<int> nums;
  for (map<int, int>::iterator it = cnt.begin(); it != cnt.end(); it++) {
    nums.push_back(it->first);
  }
  sort(nums.begin(), nums.end());
  vector<int> freq;
  for (int i = 0; i < nums.size(); i++) {
    freq.push_back(cnt[nums[i]]);
  }
  memset(dp, 0, sizeof(dp));
  int r = nums.size();
  long long ans = 0;
  long long multi = l / n;
  for (int i = 0; i < nums.size(); i++) {
    dp[r + i] = freq[i];
    ans += (dp[r + i] * multi) % 1000000007;
    ans %= 1000000007;
  }
  for (int x = 2; x <= min(multi, (long long)k); x++) {
    long long run = 0;
    for (int i = 0; i < nums.size(); i++) {
      run += dp[r * (x - 1) + i];
      dp[r * x + i] = (run * freq[i]) % 1000000007;
      ans += (dp[r * x + i] * ((multi - x + 1) % 1000000007)) % 1000000007;
      ans %= 1000000007;
    }
  }
  for (int i = 0; i < nums.size(); i++) {
    id[nums[i]] = i;
  }
  vector<int> freq2;
  for (int i = 0; i < nums.size(); i++) {
    freq2.push_back(0);
  }
  long long left = l % n;
  for (int i = 0; i < left; i++) {
    freq2[id[a[i]]]++;
  }
  ans += left;
  for (int i = 0; i < nums.size(); i++) {
    long long sum = 0;
    for (int x = 1; x < k; x++) {
      sum += dp[x * r + i];
      sum %= 1000000007;
    }
    ans += (sum * left) % 1000000007;
    ans %= 1000000007;
    left -= freq2[i];
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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;
long long l;
int a[1000000];
long long dp[2000000];
map<int, int> cnt, id;
int main() {
  scanf("%d %I64d %d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    cnt[a[i]]++;
  }
  if (k == 1) {
    printf("%I64d\n", l % 1000000007);
    return 0;
  }
  if (l <= n) {
    printf("%I64d\n", l);
    return 0;
  }
  vector<int> nums;
  for (map<int, int>::iterator it = cnt.begin(); it != cnt.end(); it++) {
    nums.push_back(it->first);
  }
  sort(nums.begin(), nums.end());
  vector<int> freq;
  for (int i = 0; i < nums.size(); i++) {
    freq.push_back(cnt[nums[i]]);
  }
  memset(dp, 0, sizeof(dp));
  int r = nums.size();
  long long ans = 0;
  long long multi = l / n;
  for (int i = 0; i < nums.size(); i++) {
    dp[r + i] = freq[i];
    ans += (dp[r + i] * multi) % 1000000007;
    ans %= 1000000007;
  }
  for (int x = 2; x <= min(multi, (long long)k); x++) {
    long long run = 0;
    for (int i = 0; i < nums.size(); i++) {
      run += dp[r * (x - 1) + i];
      dp[r * x + i] = (run * freq[i]) % 1000000007;
      ans += (dp[r * x + i] * ((multi - x + 1) % 1000000007)) % 1000000007;
      ans %= 1000000007;
    }
  }
  for (int i = 0; i < nums.size(); i++) {
    id[nums[i]] = i;
  }
  vector<int> freq2;
  for (int i = 0; i < nums.size(); i++) {
    freq2.push_back(0);
  }
  long long left = l % n;
  for (int i = 0; i < left; i++) {
    freq2[id[a[i]]]++;
  }
  ans += left;
  for (int i = 0; i < nums.size(); i++) {
    long long sum = 0;
    for (int x = 1; x < k; x++) {
      sum += dp[x * r + i];
      sum %= 1000000007;
    }
    ans += (sum * left) % 1000000007;
    ans %= 1000000007;
    left -= freq2[i];
  }
  printf("%I64d\n", ans);
  return 0;
}

```