#include <bits/stdc++.h>
using namespace std;
const unsigned long long int MOD = 1000000007;
int n, m;
unsigned long long int l;
unsigned long long int** dp;
unsigned long long int** sum;
int* mass;
int* tail;
int* check;
void sum_back(int j) {
  for (int i = n; i > 1; --i) {
    if (mass[i] == mass[i - 1]) {
      sum[i - 1][j] = sum[i][j];
    }
  }
}
void update(int j) {
  for (int i = 1; i <= n; ++i) {
    dp[i][j] = sum[i][j - 1];
    sum[i][j] = (sum[i - 1][j] + dp[i][j]) % MOD;
  }
  sum_back(j);
}
int main() {
  cin >> n >> l >> m;
  if (n > l) n = l;
  mass = new int[n + 5];
  tail = new int[n + 5];
  check = new int[n + 5];
  unsigned long long int* m1 = new unsigned long long int[(n + 5) * (m + 5)];
  unsigned long long int* m2 = new unsigned long long int[(n + 5) * (m + 5)];
  dp = new unsigned long long int*[n + 5];
  sum = new unsigned long long int*[n + 5];
  for (int i = 0; i < n + 5; ++i) {
    dp[i] = &m1[i * (m + 5)];
    sum[i] = &m2[i * (m + 5)];
  }
  for (int i = 0; i < n + 5; ++i) {
    for (int j = 0; j < m + 5; ++j) {
      dp[i][j] = 0;
      sum[i][j] = 0;
    }
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", mass + i);
    check[i] = 0;
  }
  unsigned long long int x = l % n;
  unsigned long long int mp = l / n;
  for (int i = 1; i <= x; ++i) {
    tail[i] = mass[i];
  }
  sort(mass + 1, mass + n + 1);
  sort(tail + 1, tail + x + 1);
  int num = 1;
  for (int i = 1; i <= x; ++i) {
    while (tail[i] != mass[num]) {
      ++num;
    }
    check[num] = 1;
    ++num;
  }
  for (int i = 1; i <= n; ++i) {
    dp[i][0] = 1;
    sum[i][0] = i;
  }
  sum_back(0);
  for (int j = 1; j < m; ++j) {
    update(j);
  }
  unsigned long long int ans = 0;
  for (int i = 0; i < n + 5; ++i) {
    for (int j = 0; j < m + 5; ++j) {
      ans += (mp + check[i] > j) ? ((mp + check[i] - j) % MOD) * dp[i][j] : 0;
      ans = ans % MOD;
    }
  }
  cout << ans << "\n";
}

### Prompt

Create a solution in cpp for 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;
const unsigned long long int MOD = 1000000007;
int n, m;
unsigned long long int l;
unsigned long long int** dp;
unsigned long long int** sum;
int* mass;
int* tail;
int* check;
void sum_back(int j) {
  for (int i = n; i > 1; --i) {
    if (mass[i] == mass[i - 1]) {
      sum[i - 1][j] = sum[i][j];
    }
  }
}
void update(int j) {
  for (int i = 1; i <= n; ++i) {
    dp[i][j] = sum[i][j - 1];
    sum[i][j] = (sum[i - 1][j] + dp[i][j]) % MOD;
  }
  sum_back(j);
}
int main() {
  cin >> n >> l >> m;
  if (n > l) n = l;
  mass = new int[n + 5];
  tail = new int[n + 5];
  check = new int[n + 5];
  unsigned long long int* m1 = new unsigned long long int[(n + 5) * (m + 5)];
  unsigned long long int* m2 = new unsigned long long int[(n + 5) * (m + 5)];
  dp = new unsigned long long int*[n + 5];
  sum = new unsigned long long int*[n + 5];
  for (int i = 0; i < n + 5; ++i) {
    dp[i] = &m1[i * (m + 5)];
    sum[i] = &m2[i * (m + 5)];
  }
  for (int i = 0; i < n + 5; ++i) {
    for (int j = 0; j < m + 5; ++j) {
      dp[i][j] = 0;
      sum[i][j] = 0;
    }
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", mass + i);
    check[i] = 0;
  }
  unsigned long long int x = l % n;
  unsigned long long int mp = l / n;
  for (int i = 1; i <= x; ++i) {
    tail[i] = mass[i];
  }
  sort(mass + 1, mass + n + 1);
  sort(tail + 1, tail + x + 1);
  int num = 1;
  for (int i = 1; i <= x; ++i) {
    while (tail[i] != mass[num]) {
      ++num;
    }
    check[num] = 1;
    ++num;
  }
  for (int i = 1; i <= n; ++i) {
    dp[i][0] = 1;
    sum[i][0] = i;
  }
  sum_back(0);
  for (int j = 1; j < m; ++j) {
    update(j);
  }
  unsigned long long int ans = 0;
  for (int i = 0; i < n + 5; ++i) {
    for (int j = 0; j < m + 5; ++j) {
      ans += (mp + check[i] > j) ? ((mp + check[i] - j) % MOD) * dp[i][j] : 0;
      ans = ans % MOD;
    }
  }
  cout << ans << "\n";
}

```