#include <bits/stdc++.h>
using namespace std;
const int mm = 1000000007;
long long len, c, w;
int now, pre, i, j, k, n, m, f[2][1100001], a[1100001], b[1100001], p[1100001],
    tmp, ans;
bool cmp(int x, int y) { return a[x] < a[y]; }
int main() {
  scanf("%d%I64d%d", &n, &len, &m);
  w = len / n;
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  now = 0;
  pre = 1;
  for (i = 1, b[0] = 0; i <= n; i++) f[now][i] = 1, b[++b[0]] = 1, p[i] = i;
  sort(p + 1, p + 1 + n, cmp);
  for (i = 2; i <= m; i++) {
    now ^= 1;
    pre ^= 1;
    for (j = 1, k = tmp = 0; j <= n; j++) {
      for (; k < n && a[p[k + 1]] <= a[p[j]];)
        tmp = (tmp + f[pre][p[++k]]) % mm;
      f[now][p[j]] = tmp;
    }
    for (j = 1; j <= n; j++) b[++b[0]] = f[now][j];
  }
  for (i = 1, k = ans = 0; i <= m; i++)
    for (j = 1; j <= n; j++) {
      tmp = b[++k];
      c = w + ((j <= len % n) ? 1 : 0);
      if (c < i) continue;
      ans = (ans + (long long)((c - i + 1) % mm) * (long long)tmp % mm) % mm;
    }
  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 mm = 1000000007;
long long len, c, w;
int now, pre, i, j, k, n, m, f[2][1100001], a[1100001], b[1100001], p[1100001],
    tmp, ans;
bool cmp(int x, int y) { return a[x] < a[y]; }
int main() {
  scanf("%d%I64d%d", &n, &len, &m);
  w = len / n;
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  now = 0;
  pre = 1;
  for (i = 1, b[0] = 0; i <= n; i++) f[now][i] = 1, b[++b[0]] = 1, p[i] = i;
  sort(p + 1, p + 1 + n, cmp);
  for (i = 2; i <= m; i++) {
    now ^= 1;
    pre ^= 1;
    for (j = 1, k = tmp = 0; j <= n; j++) {
      for (; k < n && a[p[k + 1]] <= a[p[j]];)
        tmp = (tmp + f[pre][p[++k]]) % mm;
      f[now][p[j]] = tmp;
    }
    for (j = 1; j <= n; j++) b[++b[0]] = f[now][j];
  }
  for (i = 1, k = ans = 0; i <= m; i++)
    for (j = 1; j <= n; j++) {
      tmp = b[++k];
      c = w + ((j <= len % n) ? 1 : 0);
      if (c < i) continue;
      ans = (ans + (long long)((c - i + 1) % mm) * (long long)tmp % mm) % mm;
    }
  printf("%d\n", ans);
  return 0;
}

```