#include <bits/stdc++.h>
using namespace std;
int N, K;
long long L;
int A[1000000];
int T[1000005];
void init() {
  for (int i = 0; i <= N; i++) T[i] = 0;
}
void update(int i, int d) {
  for (i++; i <= N; i += (i & -i)) T[i] = (T[i] + d) % 1000000007;
}
int getSum(int i) {
  int c = 0;
  for (i++; i > 0; i -= (i & -i)) c = (c + T[i]) % 1000000007;
  return c;
}
long long dp[1000000];
long long occ[1000000];
int main() {
  scanf("%d %I64d %d", &N, &L, &K);
  vector<int> lst;
  for (int i = 0; i < N; i++) {
    scanf("%d", &A[i]);
    lst.push_back(A[i]);
  }
  sort(lst.begin(), lst.end());
  lst.resize(distance(lst.begin(), unique(lst.begin(), lst.end())));
  for (int i = 0; i < N; i++) {
    A[i] = lower_bound(lst.begin(), lst.end(), A[i]) - lst.begin();
  }
  for (int i = 0; i < N; i++) {
    if (i < L)
      occ[i] = (L - 1 - i) / N + 1;
    else
      occ[i] = 0;
  }
  long long ans = 0;
  for (int i = 0; i < N; i++) {
    dp[i] = 1;
    ans = (ans + dp[i] * (max(0LL, occ[i] - 0) % 1000000007)) % 1000000007;
  }
  for (int x = 2; x <= K; x++) {
    init();
    for (int i = 0; i < N; i++) update(A[i], dp[i]);
    for (int i = 0; i < N; i++) {
      dp[i] = getSum(A[i]);
      ans = (ans + dp[i] * (max(0LL, occ[i] - (x - 1)) % 1000000007)) %
            1000000007;
    }
  }
  printf("%d\n", (int)ans);
}

### Prompt

Develop a solution in CPP 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>
using namespace std;
int N, K;
long long L;
int A[1000000];
int T[1000005];
void init() {
  for (int i = 0; i <= N; i++) T[i] = 0;
}
void update(int i, int d) {
  for (i++; i <= N; i += (i & -i)) T[i] = (T[i] + d) % 1000000007;
}
int getSum(int i) {
  int c = 0;
  for (i++; i > 0; i -= (i & -i)) c = (c + T[i]) % 1000000007;
  return c;
}
long long dp[1000000];
long long occ[1000000];
int main() {
  scanf("%d %I64d %d", &N, &L, &K);
  vector<int> lst;
  for (int i = 0; i < N; i++) {
    scanf("%d", &A[i]);
    lst.push_back(A[i]);
  }
  sort(lst.begin(), lst.end());
  lst.resize(distance(lst.begin(), unique(lst.begin(), lst.end())));
  for (int i = 0; i < N; i++) {
    A[i] = lower_bound(lst.begin(), lst.end(), A[i]) - lst.begin();
  }
  for (int i = 0; i < N; i++) {
    if (i < L)
      occ[i] = (L - 1 - i) / N + 1;
    else
      occ[i] = 0;
  }
  long long ans = 0;
  for (int i = 0; i < N; i++) {
    dp[i] = 1;
    ans = (ans + dp[i] * (max(0LL, occ[i] - 0) % 1000000007)) % 1000000007;
  }
  for (int x = 2; x <= K; x++) {
    init();
    for (int i = 0; i < N; i++) update(A[i], dp[i]);
    for (int i = 0; i < N; i++) {
      dp[i] = getSum(A[i]);
      ans = (ans + dp[i] * (max(0LL, occ[i] - (x - 1)) % 1000000007)) %
            1000000007;
    }
  }
  printf("%d\n", (int)ans);
}

```