#include <bits/stdc++.h>
using namespace std;
void gt(int &a, int b) {
  if (a < b) a = b;
}
string n2s(int a) {
  stringstream ss;
  ss << a;
  return ss.str();
}
string getKey(int v, int h) { return n2s(v) + "-" + n2s(h); }
int a[1000005], b[1000005];
long long g[1000005]{0};
set<int> s;
long long mul(long long x, long long y) {
  return ((x % 1000000007) * (y % 1000000007)) % 1000000007;
}
long long add(long long x, long long y) {
  return ((x % 1000000007) + (y % 1000000007)) % 1000000007;
}
int main() {
  long long l;
  int m = -1, n, k;
  cin >> n >> l >> k;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
    g[i] = 1;
    m = max(m, a[i]);
  }
  long long ans = l % 1000000007, u = l / n, v = l % n;
  sort(b + 1, b + n + 1);
  for (int i = 2; i <= k; i++) {
    for (int j = 1; j <= n; j++) {
      int l = lower_bound(b + 1, b + n, b[j]) - b;
      int r = b[j] == m ? n + 1 : upper_bound(b + 1, b + n, b[j]) - b;
      g[j] = add(g[l - 1], mul(g[j], r - l));
    }
    if (u >= i)
      for (int j = 1; j <= n; j++) ans = add(ans, mul(g[j], u - i + 1));
    if (v > 0 && u + 1 >= i)
      for (int j = 1; j <= v; j++) {
        int t = lower_bound(b + 1, b + n, a[j]) - b;
        ans = add(ans, g[t]);
      }
  }
  cout << ans << endl;
  return 0;
}

### 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;
void gt(int &a, int b) {
  if (a < b) a = b;
}
string n2s(int a) {
  stringstream ss;
  ss << a;
  return ss.str();
}
string getKey(int v, int h) { return n2s(v) + "-" + n2s(h); }
int a[1000005], b[1000005];
long long g[1000005]{0};
set<int> s;
long long mul(long long x, long long y) {
  return ((x % 1000000007) * (y % 1000000007)) % 1000000007;
}
long long add(long long x, long long y) {
  return ((x % 1000000007) + (y % 1000000007)) % 1000000007;
}
int main() {
  long long l;
  int m = -1, n, k;
  cin >> n >> l >> k;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
    g[i] = 1;
    m = max(m, a[i]);
  }
  long long ans = l % 1000000007, u = l / n, v = l % n;
  sort(b + 1, b + n + 1);
  for (int i = 2; i <= k; i++) {
    for (int j = 1; j <= n; j++) {
      int l = lower_bound(b + 1, b + n, b[j]) - b;
      int r = b[j] == m ? n + 1 : upper_bound(b + 1, b + n, b[j]) - b;
      g[j] = add(g[l - 1], mul(g[j], r - l));
    }
    if (u >= i)
      for (int j = 1; j <= n; j++) ans = add(ans, mul(g[j], u - i + 1));
    if (v > 0 && u + 1 >= i)
      for (int j = 1; j <= v; j++) {
        int t = lower_bound(b + 1, b + n, a[j]) - b;
        ans = add(ans, g[t]);
      }
  }
  cout << ans << endl;
  return 0;
}

```