#include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(T &t) {
  t = 0;
  char ch = getchar();
  int f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  do {
    (t *= 10) += ch - '0';
    ch = getchar();
  } while ('0' <= ch && ch <= '9');
  t *= f;
}
const int maxn = (1e6) + 10;
const int mod = (1e9) + 7;
int n, k, ans, dp[maxn], a[maxn], s[maxn];
long long len;
pair<int, int> p[maxn];
void update(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int main() {
  read(n), read(len), read(k);
  for (int i = 0; i < n; i++) read(a[i]);
  for (int i = 0; i < n * k; i++) p[i] = make_pair(a[i % n], i);
  sort(p, p + n * k);
  for (int i = 0; i < n * k; i++) {
    if (p[i].second < n)
      dp[p[i].second] = 1;
    else
      dp[p[i].second] = s[p[i].second / n - 1];
    update(s[p[i].second / n], dp[p[i].second]);
    if (p[i].second < len)
      update(ans,
             ((len - p[i].second - 1) / n + 1) % mod * dp[p[i].second] % mod);
  }
  printf("%d\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;
template <typename T>
void read(T &t) {
  t = 0;
  char ch = getchar();
  int f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  do {
    (t *= 10) += ch - '0';
    ch = getchar();
  } while ('0' <= ch && ch <= '9');
  t *= f;
}
const int maxn = (1e6) + 10;
const int mod = (1e9) + 7;
int n, k, ans, dp[maxn], a[maxn], s[maxn];
long long len;
pair<int, int> p[maxn];
void update(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int main() {
  read(n), read(len), read(k);
  for (int i = 0; i < n; i++) read(a[i]);
  for (int i = 0; i < n * k; i++) p[i] = make_pair(a[i % n], i);
  sort(p, p + n * k);
  for (int i = 0; i < n * k; i++) {
    if (p[i].second < n)
      dp[p[i].second] = 1;
    else
      dp[p[i].second] = s[p[i].second / n - 1];
    update(s[p[i].second / n], dp[p[i].second]);
    if (p[i].second < len)
      update(ans,
             ((len - p[i].second - 1) / n + 1) % mod * dp[p[i].second] % mod);
  }
  printf("%d\n", ans);
  return 0;
}

```