#include <bits/stdc++.h>
using namespace std;
template <class T>
T sqr(T x) {
  return x * x;
}
int now, nex;
const int maxn = 1e6 + 10;
long long dp[maxn][2];
int a[maxn];
int n, k;
long long l, ans;
const int mod = 1e9 + 7;
int b[maxn];
long long x, c;
inline void add_to_answer(int i, int j) {
  if (i <= x)
    ans = (ans + dp[i][nex] * ((c - j + 1) % mod) % mod) % mod;
  else if (j <= min(c - 1, 1ll * k))
    ans = (ans + dp[i][nex] * ((c - j) % mod) % mod) % mod;
}
int main() {
  cin >> n >> l >> k;
  c = (l + n - 1) / n;
  x = (l - 1) % n;
  now = 0;
  nex = 1;
  for (int i = 0; i <= n - 1; i++) {
    scanf("%d", &a[i]);
    b[i] = i;
    dp[i][nex] = 1;
    add_to_answer(i, 1);
  }
  sort(b, b + n, [](const int &i, const int &j) {
    return (pair<int, int>(a[i], i) < pair<int, int>(a[j], j));
  });
  long long limit = min(c, 1ll * k);
  for (int i = 2; i <= limit; i++) {
    now ^= 1;
    nex ^= 1;
    int tmp = 0;
    int tmp2 = 0;
    long long sum = 0;
    for (int j = 0; j <= n - 1; j++) {
      while (tmp < n && a[b[tmp]] <= a[b[j]]) {
        sum = (sum + dp[b[tmp]][now]) % mod;
        tmp++;
      }
      dp[b[j]][nex] = sum;
      add_to_answer(b[j], i);
    }
  }
  cout << ans << endl;
}

### Prompt

Your task is to create 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;
template <class T>
T sqr(T x) {
  return x * x;
}
int now, nex;
const int maxn = 1e6 + 10;
long long dp[maxn][2];
int a[maxn];
int n, k;
long long l, ans;
const int mod = 1e9 + 7;
int b[maxn];
long long x, c;
inline void add_to_answer(int i, int j) {
  if (i <= x)
    ans = (ans + dp[i][nex] * ((c - j + 1) % mod) % mod) % mod;
  else if (j <= min(c - 1, 1ll * k))
    ans = (ans + dp[i][nex] * ((c - j) % mod) % mod) % mod;
}
int main() {
  cin >> n >> l >> k;
  c = (l + n - 1) / n;
  x = (l - 1) % n;
  now = 0;
  nex = 1;
  for (int i = 0; i <= n - 1; i++) {
    scanf("%d", &a[i]);
    b[i] = i;
    dp[i][nex] = 1;
    add_to_answer(i, 1);
  }
  sort(b, b + n, [](const int &i, const int &j) {
    return (pair<int, int>(a[i], i) < pair<int, int>(a[j], j));
  });
  long long limit = min(c, 1ll * k);
  for (int i = 2; i <= limit; i++) {
    now ^= 1;
    nex ^= 1;
    int tmp = 0;
    int tmp2 = 0;
    long long sum = 0;
    for (int j = 0; j <= n - 1; j++) {
      while (tmp < n && a[b[tmp]] <= a[b[j]]) {
        sum = (sum + dp[b[tmp]][now]) % mod;
        tmp++;
      }
      dp[b[j]][nex] = sum;
      add_to_answer(b[j], i);
    }
  }
  cout << ans << endl;
}

```