#include <bits/stdc++.h>
using namespace std;
const int P = 1000000007;
const int N = 1123456;
long long l;
int n, m, u;
int a[N], v[N], c[N], f[N];
void upd(int& x, int y) {
  x += y;
  if (x >= P) x -= P;
}
int main() {
  cin >> n >> l >> m;
  for (int i = 0; i < n; ++i) scanf("%d", a + i), v[i] = a[i];
  sort(v, v + n);
  u = unique(v, v + n) - v;
  for (int i = 0; i < n; ++i) a[i] = lower_bound(v, v + u, a[i]) - v + 1;
  long long s = l / n;
  int ans = 0;
  int ed = (int)min((long long)m, s + 1);
  for (int j = 0; j < ed; ++j) {
    for (int i = 1; i < u + 1; ++i) c[i] = 0;
    for (int i = 0; i < n; ++i)
      for (int x = a[i]; x <= u; x += x & -x) upd(c[x], f[i]);
    for (int i = 0; i < n; ++i) {
      f[i] = j == 0;
      for (int x = a[i]; x > 0; x -= x & -x) upd(f[i], c[x]);
      if (j < s) ans = (ans + (s - j) % P * f[i]) % P;
      if (ans < 0) ans += P;
      if (i < l % n) upd(ans, f[i]);
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Develop a solution in CPP 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 P = 1000000007;
const int N = 1123456;
long long l;
int n, m, u;
int a[N], v[N], c[N], f[N];
void upd(int& x, int y) {
  x += y;
  if (x >= P) x -= P;
}
int main() {
  cin >> n >> l >> m;
  for (int i = 0; i < n; ++i) scanf("%d", a + i), v[i] = a[i];
  sort(v, v + n);
  u = unique(v, v + n) - v;
  for (int i = 0; i < n; ++i) a[i] = lower_bound(v, v + u, a[i]) - v + 1;
  long long s = l / n;
  int ans = 0;
  int ed = (int)min((long long)m, s + 1);
  for (int j = 0; j < ed; ++j) {
    for (int i = 1; i < u + 1; ++i) c[i] = 0;
    for (int i = 0; i < n; ++i)
      for (int x = a[i]; x <= u; x += x & -x) upd(c[x], f[i]);
    for (int i = 0; i < n; ++i) {
      f[i] = j == 0;
      for (int x = a[i]; x > 0; x -= x & -x) upd(f[i], c[x]);
      if (j < s) ans = (ans + (s - j) % P * f[i]) % P;
      if (ans < 0) ans += P;
      if (i < l % n) upd(ans, f[i]);
    }
  }
  printf("%d\n", ans);
  return 0;
}

```