#include <bits/stdc++.h>
using namespace std;
const int N = 1048576;
const int M = 1000000007;
int n, k;
long long l;
map<int, int> id;
int a[N], tn;
int sum[N];
int f[N];
long long ans = 0;
int main(int argc, char const *argv[]) {
  cin >> n >> l >> k;
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
    id[a[i]] = 0;
  }
  for (map<int, int>::iterator i = id.begin(); i != id.end(); ++i)
    i->second = tn++;
  for (int i = 0; i < n; ++i) a[i] = id[a[i]];
  long long r = l / n;
  for (int x = 1; x <= k; ++x) {
    for (int i = 0; i < n; ++i) {
      if (x == 1)
        f[i] = 1;
      else
        f[i] = sum[a[i]];
      if (i < l % n) {
        ans = (ans + (long long)f[i] * (max(0ll, r - x + 2) % M) % M) % M;
      } else
        ans = (ans + (long long)f[i] * (max(0ll, r - x + 1) % M) % M) % M;
    }
    for (int i = 0; i < tn; ++i) sum[i] = 0;
    for (int i = 0; i < n; ++i) {
      sum[a[i]] += f[i];
      sum[a[i]] %= M;
    }
    for (int i = 1; i < tn; ++i) {
      sum[i] += sum[i - 1];
      sum[i] %= M;
    }
  }
  printf("%d\n", (int)ans);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 N = 1048576;
const int M = 1000000007;
int n, k;
long long l;
map<int, int> id;
int a[N], tn;
int sum[N];
int f[N];
long long ans = 0;
int main(int argc, char const *argv[]) {
  cin >> n >> l >> k;
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
    id[a[i]] = 0;
  }
  for (map<int, int>::iterator i = id.begin(); i != id.end(); ++i)
    i->second = tn++;
  for (int i = 0; i < n; ++i) a[i] = id[a[i]];
  long long r = l / n;
  for (int x = 1; x <= k; ++x) {
    for (int i = 0; i < n; ++i) {
      if (x == 1)
        f[i] = 1;
      else
        f[i] = sum[a[i]];
      if (i < l % n) {
        ans = (ans + (long long)f[i] * (max(0ll, r - x + 2) % M) % M) % M;
      } else
        ans = (ans + (long long)f[i] * (max(0ll, r - x + 1) % M) % M) % M;
    }
    for (int i = 0; i < tn; ++i) sum[i] = 0;
    for (int i = 0; i < n; ++i) {
      sum[a[i]] += f[i];
      sum[a[i]] %= M;
    }
    for (int i = 1; i < tn; ++i) {
      sum[i] += sum[i - 1];
      sum[i] %= M;
    }
  }
  printf("%d\n", (int)ans);
  return 0;
}

```