#include <bits/stdc++.h>
using namespace std;
long long mod = 1000LL * 1000LL * 1000LL + 7;
long long n, k, len;
vector<long long> a, b;
vector<vector<long long> > dp;
void init() {
  cin >> n >> len >> k;
  a.resize(n + 1);
  b.resize(n);
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    b[i - 1] = a[i];
  }
  sort(a.begin(), a.end());
  dp.resize(k + 1, vector<long long>(n + 1, 0));
  for (int i = 1; i <= n; ++i) {
    dp[1][i] = 1;
  }
}
void solve() {
  if (len <= n) {
    cout << len;
    return;
  }
  long long sum = 0;
  int l = 1, r = 1;
  for (int i = 2; i <= k; ++i) {
    sum = 0;
    l = 1;
    r = 1;
    while (l <= n) {
      r = l;
      while (r <= n && a[r] == a[l]) {
        sum = (sum + dp[i - 1][r]) % mod;
        ++r;
      }
      for (int j = l; j < r; ++j) {
        dp[i][j] = (dp[i][j] + sum) % mod;
      }
      l = r;
    }
  }
  long long res = 0, pos = min(len / n, k);
  for (int i = 1; i <= pos; ++i) {
    for (int j = 1; j <= n; ++j) {
      res = (res + (((len / n - i + 1) % mod) * dp[i][j]) % mod) % mod;
    }
  }
  if (len % n > 0) {
    if (len / n < k) {
      pos = len / n + 1;
      for (int i = 1; i <= len % n; ++i) {
        int d =
            std::distance(a.begin(), lower_bound(a.begin(), a.end(), b[i - 1]));
        res = (res + dp[pos][d]) % mod;
      }
    }
    pos = min(len / n, k);
    for (int i = 1; i <= pos; ++i) {
      for (int j = 1; j <= len % n; ++j) {
        int d =
            std::distance(a.begin(), lower_bound(a.begin(), a.end(), b[j - 1]));
        res = (res + dp[i][d]) % mod;
      }
    }
  }
  cout << res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  init();
  solve();
  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;
long long mod = 1000LL * 1000LL * 1000LL + 7;
long long n, k, len;
vector<long long> a, b;
vector<vector<long long> > dp;
void init() {
  cin >> n >> len >> k;
  a.resize(n + 1);
  b.resize(n);
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    b[i - 1] = a[i];
  }
  sort(a.begin(), a.end());
  dp.resize(k + 1, vector<long long>(n + 1, 0));
  for (int i = 1; i <= n; ++i) {
    dp[1][i] = 1;
  }
}
void solve() {
  if (len <= n) {
    cout << len;
    return;
  }
  long long sum = 0;
  int l = 1, r = 1;
  for (int i = 2; i <= k; ++i) {
    sum = 0;
    l = 1;
    r = 1;
    while (l <= n) {
      r = l;
      while (r <= n && a[r] == a[l]) {
        sum = (sum + dp[i - 1][r]) % mod;
        ++r;
      }
      for (int j = l; j < r; ++j) {
        dp[i][j] = (dp[i][j] + sum) % mod;
      }
      l = r;
    }
  }
  long long res = 0, pos = min(len / n, k);
  for (int i = 1; i <= pos; ++i) {
    for (int j = 1; j <= n; ++j) {
      res = (res + (((len / n - i + 1) % mod) * dp[i][j]) % mod) % mod;
    }
  }
  if (len % n > 0) {
    if (len / n < k) {
      pos = len / n + 1;
      for (int i = 1; i <= len % n; ++i) {
        int d =
            std::distance(a.begin(), lower_bound(a.begin(), a.end(), b[i - 1]));
        res = (res + dp[pos][d]) % mod;
      }
    }
    pos = min(len / n, k);
    for (int i = 1; i <= pos; ++i) {
      for (int j = 1; j <= len % n; ++j) {
        int d =
            std::distance(a.begin(), lower_bound(a.begin(), a.end(), b[j - 1]));
        res = (res + dp[i][d]) % mod;
      }
    }
  }
  cout << res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  init();
  solve();
  return 0;
}

```