#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 1e6 + 10, MOD = 1e9 + 7;
int *D[MAX_N];
int *F[MAX_N];
pair<int, int> A[MAX_N];
int P[MAX_N];
inline int add(int a, int b) {
  a += b;
  return a >= MOD ? a - MOD : a;
}
int main() {
  int N, K;
  long long L;
  scanf(
      "%d"
      "%lld"
      "%d",
      &N, &L, &K);
  for (int i = 0; i < (N); ++i) scanf("%d", &A[i].first), A[i].second = i;
  sort(A, A + N);
  for (int i = (N)-1; i >= 0; --i)
    if (i < N - 1 && A[i].first == A[i + 1].first)
      P[i] = P[i + 1];
    else
      P[i] = i;
  for (int i = 0; i < (K); ++i) {
    D[i] = new int[N + 3]();
    F[i] = new int[N + 3]();
    for (int j = 0; j < (N); ++j) {
      D[i][j] = add((i ? F[i - 1][P[j] + 1] : 0), 1);
      F[i][j + 1] = add(F[i][j], D[i][j]);
    }
  }
  if (L <= N * K) {
    int ans = 0;
    for (int i = 0; i < (K); ++i)
      for (int j = 0; j < (N); ++j)
        if (i * N + A[j].second < L) ans = add(ans, D[i][j]);
    printf("%d\n", ans);
  } else {
    int ans1 = 0, ans2 = 0;
    long long cnt = (L - N * (K - 1)) / N;
    for (int i = 0; i < (K - 1); ++i)
      for (int j = 0; j < (N); ++j) ans1 = add(ans1, D[i][j]);
    for (int j = 0; j < (N); ++j) ans2 = add(ans2, D[K - 1][j]);
    ans1 = int((ans1 + (cnt % MOD) * (ans2 % MOD)) % MOD);
    L %= N;
    for (int j = 0; j < (N); ++j)
      if (A[j].second < L) {
        ans1 = add(ans1, D[K - 1][j]);
      }
    printf("%d\n", ans1);
  }
  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;
const int MAX_N = 1e6 + 10, MOD = 1e9 + 7;
int *D[MAX_N];
int *F[MAX_N];
pair<int, int> A[MAX_N];
int P[MAX_N];
inline int add(int a, int b) {
  a += b;
  return a >= MOD ? a - MOD : a;
}
int main() {
  int N, K;
  long long L;
  scanf(
      "%d"
      "%lld"
      "%d",
      &N, &L, &K);
  for (int i = 0; i < (N); ++i) scanf("%d", &A[i].first), A[i].second = i;
  sort(A, A + N);
  for (int i = (N)-1; i >= 0; --i)
    if (i < N - 1 && A[i].first == A[i + 1].first)
      P[i] = P[i + 1];
    else
      P[i] = i;
  for (int i = 0; i < (K); ++i) {
    D[i] = new int[N + 3]();
    F[i] = new int[N + 3]();
    for (int j = 0; j < (N); ++j) {
      D[i][j] = add((i ? F[i - 1][P[j] + 1] : 0), 1);
      F[i][j + 1] = add(F[i][j], D[i][j]);
    }
  }
  if (L <= N * K) {
    int ans = 0;
    for (int i = 0; i < (K); ++i)
      for (int j = 0; j < (N); ++j)
        if (i * N + A[j].second < L) ans = add(ans, D[i][j]);
    printf("%d\n", ans);
  } else {
    int ans1 = 0, ans2 = 0;
    long long cnt = (L - N * (K - 1)) / N;
    for (int i = 0; i < (K - 1); ++i)
      for (int j = 0; j < (N); ++j) ans1 = add(ans1, D[i][j]);
    for (int j = 0; j < (N); ++j) ans2 = add(ans2, D[K - 1][j]);
    ans1 = int((ans1 + (cnt % MOD) * (ans2 % MOD)) % MOD);
    L %= N;
    for (int j = 0; j < (N); ++j)
      if (A[j].second < L) {
        ans1 = add(ans1, D[K - 1][j]);
      }
    printf("%d\n", ans1);
  }
  return 0;
}

```