#include <bits/stdc++.h>
using namespace std;
int A[1000000];
int S[1000000];
int T[1000000];
int P[1000000];
long long N, K, L;
long long LN;
long long R = 0;
int TL = 0;
long long mul(long long a, long long b) { return (a * b) % 1000000007; }
long long add(long long a, long long b) { return (a + b) % 1000000007; }
void addsol(long long x) { R = (R + x) % 1000000007; }
void addsol(long long x, int k) {
  long long t = LN - k + 1;
  if (t > 0) addsol(mul(x, t % 1000000007));
}
int main() {
  scanf("%I64d%I64d%I64d", &N, &L, &K);
  LN = L / N;
  K = min(K, LN + 3);
  for (int i = 0; i < N; i++) {
    scanf("%d", &A[i]);
    S[i] = A[i];
  }
  sort(S, S + N);
  TL = 0;
  T[0] = 1;
  for (int i = 1; i < N; i++) {
    if (S[i] == S[i - 1]) {
      T[TL]++;
    } else {
      P[TL] += T[TL];
      TL++;
      P[TL] = P[TL - 1];
      T[TL]++;
    }
  }
  P[TL] += T[TL];
  TL++;
  long long D[TL][K];
  long long PD[TL][K];
  PD[0][0] = 0;
  for (int i = 0; i < TL; i++) {
    D[i][0] = T[i];
    if (i != 0)
      PD[i][0] = add(D[i][0], PD[i - 1][0]);
    else
      PD[i][0] = D[i][0];
    addsol(D[i][0], 1);
  }
  for (int j = 1; j < K; j++) {
    for (int i = 0; i < TL; i++) {
      D[i][j] = mul(PD[i][j - 1], T[i]);
      if (i != 0) {
        PD[i][j] = add(PD[i - 1][j], D[i][j]);
      } else
        PD[0][j] = D[i][j];
      addsol(D[i][j], j + 1);
    }
  }
  long long rem = L % N;
  addsol(rem);
  for (int i = 0; i < rem; i++) {
    int si = (lower_bound(S, S + N, A[i]) - S);
    int di = (upper_bound(P, P + TL, si) - P);
    for (int j = 1; j < min(K, LN + 1); j++) {
      addsol(PD[di][j - 1]);
    }
  }
  cout << R << endl;
}

### Prompt

Generate a Cpp solution to 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;
int A[1000000];
int S[1000000];
int T[1000000];
int P[1000000];
long long N, K, L;
long long LN;
long long R = 0;
int TL = 0;
long long mul(long long a, long long b) { return (a * b) % 1000000007; }
long long add(long long a, long long b) { return (a + b) % 1000000007; }
void addsol(long long x) { R = (R + x) % 1000000007; }
void addsol(long long x, int k) {
  long long t = LN - k + 1;
  if (t > 0) addsol(mul(x, t % 1000000007));
}
int main() {
  scanf("%I64d%I64d%I64d", &N, &L, &K);
  LN = L / N;
  K = min(K, LN + 3);
  for (int i = 0; i < N; i++) {
    scanf("%d", &A[i]);
    S[i] = A[i];
  }
  sort(S, S + N);
  TL = 0;
  T[0] = 1;
  for (int i = 1; i < N; i++) {
    if (S[i] == S[i - 1]) {
      T[TL]++;
    } else {
      P[TL] += T[TL];
      TL++;
      P[TL] = P[TL - 1];
      T[TL]++;
    }
  }
  P[TL] += T[TL];
  TL++;
  long long D[TL][K];
  long long PD[TL][K];
  PD[0][0] = 0;
  for (int i = 0; i < TL; i++) {
    D[i][0] = T[i];
    if (i != 0)
      PD[i][0] = add(D[i][0], PD[i - 1][0]);
    else
      PD[i][0] = D[i][0];
    addsol(D[i][0], 1);
  }
  for (int j = 1; j < K; j++) {
    for (int i = 0; i < TL; i++) {
      D[i][j] = mul(PD[i][j - 1], T[i]);
      if (i != 0) {
        PD[i][j] = add(PD[i - 1][j], D[i][j]);
      } else
        PD[0][j] = D[i][j];
      addsol(D[i][j], j + 1);
    }
  }
  long long rem = L % N;
  addsol(rem);
  for (int i = 0; i < rem; i++) {
    int si = (lower_bound(S, S + N, A[i]) - S);
    int di = (upper_bound(P, P + TL, si) - P);
    for (int j = 1; j < min(K, LN + 1); j++) {
      addsol(PD[di][j - 1]);
    }
  }
  cout << R << endl;
}

```