#include <bits/stdc++.h>
using namespace std;
const int mxn = 1e6 + 6, md = 1e9 + 7;
int n, k;
long long l, ans, a[mxn], dp[mxn], s[mxn];
pair<int, int> p[mxn];
inline void add(long long& x, long long y) {
  x += y;
  if (x >= md) x -= md;
}
int main() {
  scanf("%d%lld%d", &n, &l, &k);
  for (int i = 0; i < n; ++i) scanf("%lld", a + i);
  for (int i = 0; i < n * k; ++i) p[i] = {a[i % n], i};
  sort(p, p + n * k);
  for (int i = 0; i < n * k; ++i) {
    dp[p[i].second] = (p[i].second < n) ? 1 : s[p[i].second / n - 1];
    add(s[p[i].second / n], dp[p[i].second]);
    if (p[i].second < l)
      add(ans, ((l - p[i].second - 1) / n + 1) % md * dp[p[i].second] % md);
  }
  printf("%lld\n", ans);
}

### Prompt

Create a solution in cpp for 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 mxn = 1e6 + 6, md = 1e9 + 7;
int n, k;
long long l, ans, a[mxn], dp[mxn], s[mxn];
pair<int, int> p[mxn];
inline void add(long long& x, long long y) {
  x += y;
  if (x >= md) x -= md;
}
int main() {
  scanf("%d%lld%d", &n, &l, &k);
  for (int i = 0; i < n; ++i) scanf("%lld", a + i);
  for (int i = 0; i < n * k; ++i) p[i] = {a[i % n], i};
  sort(p, p + n * k);
  for (int i = 0; i < n * k; ++i) {
    dp[p[i].second] = (p[i].second < n) ? 1 : s[p[i].second / n - 1];
    add(s[p[i].second / n], dp[p[i].second]);
    if (p[i].second < l)
      add(ans, ((l - p[i].second - 1) / n + 1) % md * dp[p[i].second] % md);
  }
  printf("%lld\n", ans);
}

```