#include <bits/stdc++.h>
using namespace std;
long long a[1000010], f[1000010], p[1000010], mod = 1000000007;
struct node {
  long long x, k;
} K[1000010];
int cmp(const void *xx, const void *yy) {
  node x = *(node *)xx, y = *(node *)yy;
  if (x.x > y.x) return 1;
  return -1;
}
int main() {
  long long n, k, i, j, o, ll, sum, last;
  long long l, ans = 0, Sum = 0, ss;
  scanf("%I64d%I64d%I64d", &n, &l, &k);
  for (i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    K[i].x = a[i];
    K[i].k = i;
  }
  qsort(K + 1, n, sizeof(node), cmp);
  o = l / n;
  l %= n;
  ll = o < k ? o : k;
  for (i = 1; i <= n; i++) f[i] = 1;
  ss = 1;
  for (i = 1; i <= ll; i++) {
    sum = 0;
    last = 1;
    for (j = 1; j <= n; j++) {
      ans =
          (ans + (f[K[j].k] * ((o - i + 1 + (K[j].k <= l)) % mod)) % mod) % mod;
      while (last <= n && K[last].x <= K[j].x)
        sum = (sum + f[K[last++].k]) % mod;
      p[K[j].k] = sum;
    }
    for (j = 1; j <= n; j++) f[j] = p[j];
    ss = (ss * n) % mod;
  }
  if (k > o)
    for (j = n; j >= 1; j--) ans = (ans + f[K[j].k] * (K[j].k <= l)) % mod;
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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;
long long a[1000010], f[1000010], p[1000010], mod = 1000000007;
struct node {
  long long x, k;
} K[1000010];
int cmp(const void *xx, const void *yy) {
  node x = *(node *)xx, y = *(node *)yy;
  if (x.x > y.x) return 1;
  return -1;
}
int main() {
  long long n, k, i, j, o, ll, sum, last;
  long long l, ans = 0, Sum = 0, ss;
  scanf("%I64d%I64d%I64d", &n, &l, &k);
  for (i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    K[i].x = a[i];
    K[i].k = i;
  }
  qsort(K + 1, n, sizeof(node), cmp);
  o = l / n;
  l %= n;
  ll = o < k ? o : k;
  for (i = 1; i <= n; i++) f[i] = 1;
  ss = 1;
  for (i = 1; i <= ll; i++) {
    sum = 0;
    last = 1;
    for (j = 1; j <= n; j++) {
      ans =
          (ans + (f[K[j].k] * ((o - i + 1 + (K[j].k <= l)) % mod)) % mod) % mod;
      while (last <= n && K[last].x <= K[j].x)
        sum = (sum + f[K[last++].k]) % mod;
      p[K[j].k] = sum;
    }
    for (j = 1; j <= n; j++) f[j] = p[j];
    ss = (ss * n) % mod;
  }
  if (k > o)
    for (j = n; j >= 1; j--) ans = (ans + f[K[j].k] * (K[j].k <= l)) % mod;
  printf("%I64d\n", ans);
  return 0;
}

```