#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
template <class T>
typename T::value_type arr_sum(const T& v, int n) {
  typename T::value_type sum = 0;
  for (int i = (0); i < (n); ++i) sum += v[i];
  return sum;
}
struct Sync_stdio {
  Sync_stdio() {
    cin.tie(NULL);
    ios_base::sync_with_stdio(false);
  }
} _sync_stdio;
int main() {
  int n, k;
  long long l;
  scanf("%d%I64d%d", &n, &l, &k);
  vector<int> a(n);
  for (int i = (0); i < (n); ++i) {
    scanf("%d", &a[i]);
  }
  vector<int> b = a;
  b.resize(l % n);
  sort(b.begin(), b.end());
  sort(a.begin(), a.end());
  vector<int> pos(n);
  for (int i = (0); i < (n); ++i) {
    pos[i] = i;
  }
  for (int i = (n - 2); i > (-1); --i) {
    if (a[i] == a[i + 1]) {
      pos[i] = pos[i + 1];
    }
  }
  vector<vector<long long>> dp(k + 1, vector<long long>(n));
  for (int i = (0); i < (n); ++i) {
    dp[0][i] = 1;
  }
  for (int i = (1); i < (k + 1); ++i) {
    for (int j = (0); j < (n); ++j) {
      dp[i][j] = dp[i - 1][pos[j]];
      dp[i][j] %= MOD;
      if (j > 0) {
        dp[i][j] += dp[i][j - 1];
        dp[i][j] %= MOD;
      }
    }
  }
  long long ans = 0;
  long long cnt1 = l / n;
  for (int i = (1); i < (k + 1); ++i) {
    if ((cnt1 - i + 1) < 0) {
      continue;
    }
    ans += ((cnt1 - i + 1) % MOD) * dp[i][n - 1];
    ans %= MOD;
  }
  long long t = l % n;
  for (int i = (0); i < (k); ++i) {
    for (int j = (0); j < (t); ++j) {
      if (cnt1 - i < 0) {
        continue;
      }
      int p = upper_bound(a.begin(), a.end(), b[j]) - a.begin() - 1;
      ans += dp[i][p];
      ans %= MOD;
    }
  }
  printf("%d", ans);
}

### 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 MOD = 1000000007;
template <class T>
typename T::value_type arr_sum(const T& v, int n) {
  typename T::value_type sum = 0;
  for (int i = (0); i < (n); ++i) sum += v[i];
  return sum;
}
struct Sync_stdio {
  Sync_stdio() {
    cin.tie(NULL);
    ios_base::sync_with_stdio(false);
  }
} _sync_stdio;
int main() {
  int n, k;
  long long l;
  scanf("%d%I64d%d", &n, &l, &k);
  vector<int> a(n);
  for (int i = (0); i < (n); ++i) {
    scanf("%d", &a[i]);
  }
  vector<int> b = a;
  b.resize(l % n);
  sort(b.begin(), b.end());
  sort(a.begin(), a.end());
  vector<int> pos(n);
  for (int i = (0); i < (n); ++i) {
    pos[i] = i;
  }
  for (int i = (n - 2); i > (-1); --i) {
    if (a[i] == a[i + 1]) {
      pos[i] = pos[i + 1];
    }
  }
  vector<vector<long long>> dp(k + 1, vector<long long>(n));
  for (int i = (0); i < (n); ++i) {
    dp[0][i] = 1;
  }
  for (int i = (1); i < (k + 1); ++i) {
    for (int j = (0); j < (n); ++j) {
      dp[i][j] = dp[i - 1][pos[j]];
      dp[i][j] %= MOD;
      if (j > 0) {
        dp[i][j] += dp[i][j - 1];
        dp[i][j] %= MOD;
      }
    }
  }
  long long ans = 0;
  long long cnt1 = l / n;
  for (int i = (1); i < (k + 1); ++i) {
    if ((cnt1 - i + 1) < 0) {
      continue;
    }
    ans += ((cnt1 - i + 1) % MOD) * dp[i][n - 1];
    ans %= MOD;
  }
  long long t = l % n;
  for (int i = (0); i < (k); ++i) {
    for (int j = (0); j < (t); ++j) {
      if (cnt1 - i < 0) {
        continue;
      }
      int p = upper_bound(a.begin(), a.end(), b[j]) - a.begin() - 1;
      ans += dp[i][p];
      ans %= MOD;
    }
  }
  printf("%d", ans);
}

```