#include <bits/stdc++.h>
using namespace std;
int dirx[] = {0, 1, 0, -1, 1, 1, -1, -1};
int diry[] = {1, 0, -1, 0, 1, -1, 1, -1};
long long powlog(long long a, long long b, long long c) {
  long long ret = 1ll;
  while (b > 0) {
    if (b % 2ll == 1ll) ret = (ret * a) % c;
    a = (a * a) % c;
    b /= 2ll;
  }
  return ret;
}
long long a[2000005];
pair<long long, int> pp[1000005];
long long b[1000005], dp[10000005];
long long ans[1000005], ans2[1000005];
int main() {
  long long n, k, r;
  long long l;
  long long m, C, ANS = 0;
  scanf("%lld%lld%lld", &n, &l, &k);
  m = l / n;
  r = l % n;
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  for (int i = 0; i < n; i++) pp[i] = make_pair(a[i], i);
  sort(pp, pp + n);
  for (int i = 0; i <= n; i++) dp[i] = 1;
  for (int i = 1; i <= k; i++) {
    C = m - i + 1;
    C %= 1000000007ll;
    for (int j = 0; j < n; j++) {
      if (i <= m + 1 && pp[j].second < r)
        ANS = (ANS + (dp[j]) % 1000000007ll) % 1000000007ll;
      if (i <= m) ANS = (ANS + (C * dp[j]) % 1000000007ll) % 1000000007ll;
    }
    b[0] = dp[0];
    for (int j = 1; j < n; j++) b[j] = (b[j - 1] + dp[j]) % 1000000007ll;
    for (int j = 0; j < n; j++) dp[j] = b[j];
    for (int j = n - 2; j >= 0; j--)
      if (pp[j].first == pp[j + 1].first) {
        dp[j] = dp[j + 1];
      }
  }
  cout << ANS << endl;
  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;
int dirx[] = {0, 1, 0, -1, 1, 1, -1, -1};
int diry[] = {1, 0, -1, 0, 1, -1, 1, -1};
long long powlog(long long a, long long b, long long c) {
  long long ret = 1ll;
  while (b > 0) {
    if (b % 2ll == 1ll) ret = (ret * a) % c;
    a = (a * a) % c;
    b /= 2ll;
  }
  return ret;
}
long long a[2000005];
pair<long long, int> pp[1000005];
long long b[1000005], dp[10000005];
long long ans[1000005], ans2[1000005];
int main() {
  long long n, k, r;
  long long l;
  long long m, C, ANS = 0;
  scanf("%lld%lld%lld", &n, &l, &k);
  m = l / n;
  r = l % n;
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  for (int i = 0; i < n; i++) pp[i] = make_pair(a[i], i);
  sort(pp, pp + n);
  for (int i = 0; i <= n; i++) dp[i] = 1;
  for (int i = 1; i <= k; i++) {
    C = m - i + 1;
    C %= 1000000007ll;
    for (int j = 0; j < n; j++) {
      if (i <= m + 1 && pp[j].second < r)
        ANS = (ANS + (dp[j]) % 1000000007ll) % 1000000007ll;
      if (i <= m) ANS = (ANS + (C * dp[j]) % 1000000007ll) % 1000000007ll;
    }
    b[0] = dp[0];
    for (int j = 1; j < n; j++) b[j] = (b[j - 1] + dp[j]) % 1000000007ll;
    for (int j = 0; j < n; j++) dp[j] = b[j];
    for (int j = n - 2; j >= 0; j--)
      if (pp[j].first == pp[j + 1].first) {
        dp[j] = dp[j + 1];
      }
  }
  cout << ANS << endl;
  return 0;
}

```