#include <bits/stdc++.h>
const int modulo = 1000000007;
int n, k;
long long l, m;
int sequence[1000000];
int __dp[1000000];
std::pair<int, int> pairs[1000000];
inline int &dp(int a, int b) { return __dp[a * n + b]; }
int main(void) {
  int i, j, x;
  int sum;
  int answer;
  scanf("%d %I64d %d", &n, &l, &k);
  m = l / n;
  if (l % n) ++m;
  if (k > m) k = m;
  if (l % n) --m;
  for (i = 0; i < n; ++i)
    scanf("%d", &sequence[i]), pairs[i] = std::make_pair(sequence[i], i);
  std::sort(pairs, pairs + n);
  for (i = 0; i < n; ++i) dp(0, i) = 1;
  for (i = 1; i < k; ++i) {
    sum = 0;
    for (j = 0; j < n; ++j) {
      if (!j || pairs[j].first != pairs[j - 1].first) {
        (sum += dp(i - 1, pairs[j].second)) %= modulo;
        x = j;
        while (++x < n && pairs[x].first == pairs[x - 1].first)
          (sum += dp(i - 1, pairs[x].second)) %= modulo;
      }
      dp(i, pairs[j].second) = sum;
    }
  }
  answer = 0;
  for (i = 0; i < l % n; ++i)
    for (j = 0; j < k && m + 1 - j > 0; ++j)
      (answer += 1ll * dp(j, i) * ((m + 1 - j) % modulo) % modulo) %= modulo;
  for (i = l % n; i < n; ++i)
    for (j = 0; j < k && m - j > 0; ++j)
      (answer += 1ll * dp(j, i) * ((m - j) % modulo) % modulo) %= modulo;
  printf("%d\n", answer);
  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>
const int modulo = 1000000007;
int n, k;
long long l, m;
int sequence[1000000];
int __dp[1000000];
std::pair<int, int> pairs[1000000];
inline int &dp(int a, int b) { return __dp[a * n + b]; }
int main(void) {
  int i, j, x;
  int sum;
  int answer;
  scanf("%d %I64d %d", &n, &l, &k);
  m = l / n;
  if (l % n) ++m;
  if (k > m) k = m;
  if (l % n) --m;
  for (i = 0; i < n; ++i)
    scanf("%d", &sequence[i]), pairs[i] = std::make_pair(sequence[i], i);
  std::sort(pairs, pairs + n);
  for (i = 0; i < n; ++i) dp(0, i) = 1;
  for (i = 1; i < k; ++i) {
    sum = 0;
    for (j = 0; j < n; ++j) {
      if (!j || pairs[j].first != pairs[j - 1].first) {
        (sum += dp(i - 1, pairs[j].second)) %= modulo;
        x = j;
        while (++x < n && pairs[x].first == pairs[x - 1].first)
          (sum += dp(i - 1, pairs[x].second)) %= modulo;
      }
      dp(i, pairs[j].second) = sum;
    }
  }
  answer = 0;
  for (i = 0; i < l % n; ++i)
    for (j = 0; j < k && m + 1 - j > 0; ++j)
      (answer += 1ll * dp(j, i) * ((m + 1 - j) % modulo) % modulo) %= modulo;
  for (i = l % n; i < n; ++i)
    for (j = 0; j < k && m - j > 0; ++j)
      (answer += 1ll * dp(j, i) * ((m - j) % modulo) % modulo) %= modulo;
  printf("%d\n", answer);
  return 0;
}

```