#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 6;
const int MAXD = 1e6 + 5;
const int mod = 1e9 + 7;
inline void add_mod(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
int a[MAXN];
struct Node {
  int x, id;
} b[MAXN];
inline bool cmp(const Node &p, const Node &q) { return p.x < q.x; }
struct Number {
  int x, cnt, ecnt;
} c[MAXN];
vector<int> f[MAXD];
int main(void) {
  int n, d;
  long long l;
  cin >> n >> l >> d;
  int nn = l % n;
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) b[i] = (Node){a[i], i};
  sort(b + 1, b + n + 1, cmp);
  int m = 0;
  for (int i = 1, j = 1; j <= n; i = j) {
    c[++m].x = b[i].x;
    while (j <= n && b[j].x == b[i].x) {
      if (b[j].id <= nn) ++c[m].ecnt;
      ++j;
    }
    c[m].cnt = j - i;
  }
  for (int i = 0; i <= d; ++i) f[i].resize(m + 5);
  for (int i = 1; i <= m; ++i) f[0][i] = 1;
  int ans = 0;
  for (int i = 1; i <= d; ++i) {
    for (int j = 1; j <= m; ++j)
      f[i][j] = (long long)f[i - 1][j] * c[j].cnt % mod;
    for (int j = 1; j <= m; ++j) add_mod(f[i][j], f[i][j - 1]);
    if (i - 1 > l / n) continue;
    add_mod(ans, (long long)(l / n - i + 1) % mod * f[i][m] % mod);
    for (int j = 1; j <= m; ++j)
      add_mod(ans, (long long)f[i - 1][j] * c[j].ecnt % mod);
  }
  printf("%d", 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 int MAXN = 1e6 + 6;
const int MAXD = 1e6 + 5;
const int mod = 1e9 + 7;
inline void add_mod(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
int a[MAXN];
struct Node {
  int x, id;
} b[MAXN];
inline bool cmp(const Node &p, const Node &q) { return p.x < q.x; }
struct Number {
  int x, cnt, ecnt;
} c[MAXN];
vector<int> f[MAXD];
int main(void) {
  int n, d;
  long long l;
  cin >> n >> l >> d;
  int nn = l % n;
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) b[i] = (Node){a[i], i};
  sort(b + 1, b + n + 1, cmp);
  int m = 0;
  for (int i = 1, j = 1; j <= n; i = j) {
    c[++m].x = b[i].x;
    while (j <= n && b[j].x == b[i].x) {
      if (b[j].id <= nn) ++c[m].ecnt;
      ++j;
    }
    c[m].cnt = j - i;
  }
  for (int i = 0; i <= d; ++i) f[i].resize(m + 5);
  for (int i = 1; i <= m; ++i) f[0][i] = 1;
  int ans = 0;
  for (int i = 1; i <= d; ++i) {
    for (int j = 1; j <= m; ++j)
      f[i][j] = (long long)f[i - 1][j] * c[j].cnt % mod;
    for (int j = 1; j <= m; ++j) add_mod(f[i][j], f[i][j - 1]);
    if (i - 1 > l / n) continue;
    add_mod(ans, (long long)(l / n - i + 1) % mod * f[i][m] % mod);
    for (int j = 1; j <= m; ++j)
      add_mod(ans, (long long)f[i - 1][j] * c[j].ecnt % mod);
  }
  printf("%d", ans);
  return 0;
}

```