#include <bits/stdc++.h>
using namespace std;
const long long mm = 1e9 + 7;
int n, k, cnt;
int a[1005000], c[1005000];
long long ans, l, p, x, s[1005000];
vector<vector<long long> > f;
int main() {
  scanf("%d%lld%d", &n, &l, &k);
  f.resize(k + 1);
  for (int i = 0; i <= k; i++) f[i].resize(n + 1);
  p = l % n;
  x = l / n;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), c[i] = a[i];
  sort(c + 1, c + 1 + n);
  cnt = unique(c + 1, c + 1 + n) - c - 1;
  for (int i = 1; i <= n; i++) a[i] = lower_bound(c + 1, c + 1 + cnt, a[i]) - c;
  for (int i = 1; i <= n; i++) f[1][a[i]]++;
  ans = l % mm;
  for (int i = 2; i <= k; i++) {
    for (int j = 1; j <= cnt; j++) s[j] = (s[j - 1] + f[i - 1][j]) % mm;
    if (i > 2 && x - i + 2 > 0) (ans += (x - i + 2) % mm * s[cnt] % mm) %= mm;
    for (int j = 1; j <= n; j++) (f[i][a[j]] += s[a[j]]) %= mm;
    if (x - i + 2 > 0)
      for (int j = 1; j <= p; j++) (ans += s[a[j]]) %= mm;
  }
  for (int j = 1; j <= cnt; j++) s[j] = (s[j - 1] + f[k][j]) % mm;
  if (k > 1 && x - k + 1 > 0) (ans += (x - k + 1) % mm * s[cnt] % mm) %= mm;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 long long mm = 1e9 + 7;
int n, k, cnt;
int a[1005000], c[1005000];
long long ans, l, p, x, s[1005000];
vector<vector<long long> > f;
int main() {
  scanf("%d%lld%d", &n, &l, &k);
  f.resize(k + 1);
  for (int i = 0; i <= k; i++) f[i].resize(n + 1);
  p = l % n;
  x = l / n;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), c[i] = a[i];
  sort(c + 1, c + 1 + n);
  cnt = unique(c + 1, c + 1 + n) - c - 1;
  for (int i = 1; i <= n; i++) a[i] = lower_bound(c + 1, c + 1 + cnt, a[i]) - c;
  for (int i = 1; i <= n; i++) f[1][a[i]]++;
  ans = l % mm;
  for (int i = 2; i <= k; i++) {
    for (int j = 1; j <= cnt; j++) s[j] = (s[j - 1] + f[i - 1][j]) % mm;
    if (i > 2 && x - i + 2 > 0) (ans += (x - i + 2) % mm * s[cnt] % mm) %= mm;
    for (int j = 1; j <= n; j++) (f[i][a[j]] += s[a[j]]) %= mm;
    if (x - i + 2 > 0)
      for (int j = 1; j <= p; j++) (ans += s[a[j]]) %= mm;
  }
  for (int j = 1; j <= cnt; j++) s[j] = (s[j - 1] + f[k][j]) % mm;
  if (k > 1 && x - k + 1 > 0) (ans += (x - k + 1) % mm * s[cnt] % mm) %= mm;
  printf("%lld\n", ans);
  return 0;
}

```