#include <bits/stdc++.h>
using namespace std;
long long int a[2000010], dp[2][2000010], b[2000006], c[1000006];
int main(int argc, char const *argv[]) {
  long long int n, l, k, temp, ans, mod, lo, hi, mid;
  long long int start, end;
  cin >> n >> l >> k;
  mod = 1000000007;
  for (int i = 0; i < n; ++i) {
    scanf("%lld", &a[i]);
  }
  for (int i = n; i < n + l % n; ++i) {
    a[i] = a[i % n];
  }
  sort(a, a + n);
  sort(a + n, a + n + l % n);
  for (int i = 0; i < n; ++i) dp[1][i] = n - i;
  dp[1][n] = 0;
  start = 0;
  end = n;
  for (int i = 0; i < n; ++i) {
    while (a[start] < a[i]) {
      start++;
    }
    c[i] = start % n;
  }
  ans = l % mod;
  for (int i = 2; i <= min(k, l / n); ++i) {
    for (int j = 0; j <= n; ++j) dp[i % 2][j] = 0;
    for (int j = n - 1; j >= 0; --j) {
      dp[i % 2][j] = dp[1 - i % 2][c[j]];
      ans = (ans + (dp[i % 2][j] * ((l / n - (i - 1)) % mod)) % mod) % mod;
    }
    for (int j = n - 1; j >= 0; --j)
      dp[i % 2][j] = (dp[i % 2][j] + dp[i % 2][j + 1]) % mod;
  }
  if (l % n != 0) {
    start = n;
    end = n + l % n;
    for (int i = 0; i < n; ++i) {
      while (a[start] < a[i] && start != end) start++;
      b[i] = start % n;
    }
    for (int i = l % n; i < n; ++i) dp[1][i] = 0;
    for (int i = 0; i < l % n; ++i) dp[1][i] = l % n - i;
    dp[1][n] = 0;
    for (int i = 2; i <= min(k, l / n + 1); ++i) {
      for (int j = 0; j <= n; ++j) dp[i % 2][j] = 0;
      if (i == 2)
        for (int j = n - 1; j >= 0; --j) {
          dp[i % 2][j] = dp[1 - i % 2][b[j]] % mod;
          ans = (ans + dp[i % 2][j]) % mod;
        }
      else {
        for (int j = n - 1; j >= 0; --j) {
          dp[i % 2][j] = (dp[1 - i % 2][c[j]]) % mod;
          ans = (ans + dp[i % 2][j]) % mod;
        }
      }
      for (int j = n - 1; j >= 0; --j)
        dp[i % 2][j] = (dp[i % 2][j] + dp[i % 2][j + 1]) % mod;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

In cpp, your task is to solve 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 int a[2000010], dp[2][2000010], b[2000006], c[1000006];
int main(int argc, char const *argv[]) {
  long long int n, l, k, temp, ans, mod, lo, hi, mid;
  long long int start, end;
  cin >> n >> l >> k;
  mod = 1000000007;
  for (int i = 0; i < n; ++i) {
    scanf("%lld", &a[i]);
  }
  for (int i = n; i < n + l % n; ++i) {
    a[i] = a[i % n];
  }
  sort(a, a + n);
  sort(a + n, a + n + l % n);
  for (int i = 0; i < n; ++i) dp[1][i] = n - i;
  dp[1][n] = 0;
  start = 0;
  end = n;
  for (int i = 0; i < n; ++i) {
    while (a[start] < a[i]) {
      start++;
    }
    c[i] = start % n;
  }
  ans = l % mod;
  for (int i = 2; i <= min(k, l / n); ++i) {
    for (int j = 0; j <= n; ++j) dp[i % 2][j] = 0;
    for (int j = n - 1; j >= 0; --j) {
      dp[i % 2][j] = dp[1 - i % 2][c[j]];
      ans = (ans + (dp[i % 2][j] * ((l / n - (i - 1)) % mod)) % mod) % mod;
    }
    for (int j = n - 1; j >= 0; --j)
      dp[i % 2][j] = (dp[i % 2][j] + dp[i % 2][j + 1]) % mod;
  }
  if (l % n != 0) {
    start = n;
    end = n + l % n;
    for (int i = 0; i < n; ++i) {
      while (a[start] < a[i] && start != end) start++;
      b[i] = start % n;
    }
    for (int i = l % n; i < n; ++i) dp[1][i] = 0;
    for (int i = 0; i < l % n; ++i) dp[1][i] = l % n - i;
    dp[1][n] = 0;
    for (int i = 2; i <= min(k, l / n + 1); ++i) {
      for (int j = 0; j <= n; ++j) dp[i % 2][j] = 0;
      if (i == 2)
        for (int j = n - 1; j >= 0; --j) {
          dp[i % 2][j] = dp[1 - i % 2][b[j]] % mod;
          ans = (ans + dp[i % 2][j]) % mod;
        }
      else {
        for (int j = n - 1; j >= 0; --j) {
          dp[i % 2][j] = (dp[1 - i % 2][c[j]]) % mod;
          ans = (ans + dp[i % 2][j]) % mod;
        }
      }
      for (int j = n - 1; j >= 0; --j)
        dp[i % 2][j] = (dp[i % 2][j] + dp[i % 2][j + 1]) % mod;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```