#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 1100000;
long long dp[maxn];
long long a[maxn];
long long tmp[maxn];
long long refer[maxn];
long long cnt[maxn];
long long n, l, k;
long long cal(long long i, long long j) { return i * k + j; }
int main() {
  scanf("%I64d%I64d%I64d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &a[i]);
    tmp[i] = a[i];
  }
  sort(tmp, tmp + n);
  for (int i = 0; i < n; i++) refer[i] = lower_bound(tmp, tmp + n, a[i]) - tmp;
  for (int i = 0; i < n; i++) dp[cal(i, 1)] = 1;
  for (int j = 2; j <= k; j++) {
    for (int i = 0; i < n; i++) cnt[i] = 0;
    for (int i = 0; i < n; i++)
      cnt[refer[i]] = (cnt[refer[i]] + dp[cal(i, j - 1)]) % mod;
    for (int i = 1; i < n; i++) cnt[i] = (cnt[i] + cnt[i - 1]) % mod;
    for (int i = 0; i < n; i++) dp[cal(i, j)] = cnt[refer[i]];
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 1; j <= k; j++) {
      long long num = l / n;
      if (i < l % n) num++;
      num = num - j + 1;
      if (num > 0) ans = (ans + ((num % mod) * dp[cal(i, j)]) % mod) % mod;
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Please formulate 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;
const int mod = 1e9 + 7;
const int maxn = 1100000;
long long dp[maxn];
long long a[maxn];
long long tmp[maxn];
long long refer[maxn];
long long cnt[maxn];
long long n, l, k;
long long cal(long long i, long long j) { return i * k + j; }
int main() {
  scanf("%I64d%I64d%I64d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &a[i]);
    tmp[i] = a[i];
  }
  sort(tmp, tmp + n);
  for (int i = 0; i < n; i++) refer[i] = lower_bound(tmp, tmp + n, a[i]) - tmp;
  for (int i = 0; i < n; i++) dp[cal(i, 1)] = 1;
  for (int j = 2; j <= k; j++) {
    for (int i = 0; i < n; i++) cnt[i] = 0;
    for (int i = 0; i < n; i++)
      cnt[refer[i]] = (cnt[refer[i]] + dp[cal(i, j - 1)]) % mod;
    for (int i = 1; i < n; i++) cnt[i] = (cnt[i] + cnt[i - 1]) % mod;
    for (int i = 0; i < n; i++) dp[cal(i, j)] = cnt[refer[i]];
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 1; j <= k; j++) {
      long long num = l / n;
      if (i < l % n) num++;
      num = num - j + 1;
      if (num > 0) ans = (ans + ((num % mod) * dp[cal(i, j)]) % mod) % mod;
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

```