#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const int N = 1e6 + 10;
long long n, k, L, a[N], rank_[N], num[N], sum[N], sum2[N];
vector<long long> d[N];
vector<pair<long long, int> > st;
int main() {
  cin >> n >> L >> k;
  for (int i = 0; i < (int)k + 1; i++) {
    for (int j = 0; j < (int)n + 1; j++) d[i].push_back(0);
  }
  for (int i = 0; i < (int)n; i++)
    scanf("%I64d", &a[i]), st.push_back(pair<long long, int>(a[i], i));
  sort(st.begin(), st.end());
  int cnt = 0, cc = 0;
  for (int i = 0; i < (int)n; i++) {
    if (!i || st[i - 1].first != st[i].first) {
      num[cnt] = cc;
      cnt++;
      cc = 1;
    } else {
      cc++;
    }
    rank_[st[i].second] = cnt;
  }
  num[cnt] = cc;
  for (int i = 0; i <= k; i++)
    for (int j = 1; j <= cnt; j++) {
      if (i == 0) {
        d[i][j] = 1;
        continue;
      }
      d[i][j] = (d[i][j] + num[j] * d[i - 1][j]) % mod;
      if (j != 1) d[i][j] = (d[i][j] + d[i][j - 1]) % mod;
    }
  long long all = L / n, yu = L % n, ans = 0;
  for (int i = 1; i <= k; i++) {
    sum[i] = sum2[i] = 0;
    for (int j = 0; j < n; j++) sum[i] = (sum[i] + d[i - 1][rank_[j]]) % mod;
    for (int j = 0; j < yu; j++) sum2[i] = (sum2[i] + d[i - 1][rank_[j]]) % mod;
  }
  for (int i = 1; i <= k; i++) {
    ans = (ans + sum[i] * (max((long long)0, all - i + 1) % mod) % mod) % mod;
    if (yu && all + 1 >= i) ans = (ans + sum2[i]) % 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 long long mod = 1e9 + 7;
const int N = 1e6 + 10;
long long n, k, L, a[N], rank_[N], num[N], sum[N], sum2[N];
vector<long long> d[N];
vector<pair<long long, int> > st;
int main() {
  cin >> n >> L >> k;
  for (int i = 0; i < (int)k + 1; i++) {
    for (int j = 0; j < (int)n + 1; j++) d[i].push_back(0);
  }
  for (int i = 0; i < (int)n; i++)
    scanf("%I64d", &a[i]), st.push_back(pair<long long, int>(a[i], i));
  sort(st.begin(), st.end());
  int cnt = 0, cc = 0;
  for (int i = 0; i < (int)n; i++) {
    if (!i || st[i - 1].first != st[i].first) {
      num[cnt] = cc;
      cnt++;
      cc = 1;
    } else {
      cc++;
    }
    rank_[st[i].second] = cnt;
  }
  num[cnt] = cc;
  for (int i = 0; i <= k; i++)
    for (int j = 1; j <= cnt; j++) {
      if (i == 0) {
        d[i][j] = 1;
        continue;
      }
      d[i][j] = (d[i][j] + num[j] * d[i - 1][j]) % mod;
      if (j != 1) d[i][j] = (d[i][j] + d[i][j - 1]) % mod;
    }
  long long all = L / n, yu = L % n, ans = 0;
  for (int i = 1; i <= k; i++) {
    sum[i] = sum2[i] = 0;
    for (int j = 0; j < n; j++) sum[i] = (sum[i] + d[i - 1][rank_[j]]) % mod;
    for (int j = 0; j < yu; j++) sum2[i] = (sum2[i] + d[i - 1][rank_[j]]) % mod;
  }
  for (int i = 1; i <= k; i++) {
    ans = (ans + sum[i] * (max((long long)0, all - i + 1) % mod) % mod) % mod;
    if (yu && all + 1 >= i) ans = (ans + sum2[i]) % mod;
  }
  cout << ans << endl;
  return 0;
}

```