#include <bits/stdc++.h>
using namespace std;
int mul(int a, int b) {
  long long c = a;
  c *= b;
  return c % 1000000007;
}
int sum(int a, int b) {
  a += b;
  return a % 1000000007;
}
int dp[1100000 * 5], a[1100000], b[1100000], c[1100000];
int n, k;
int f(int i, int x) {
  if (i == -1) return x == 0;
  int &ret = dp[i + x * n];
  if (ret != -1) return ret;
  ret = 0;
  if (x) ret = mul(c[i], f(i, x - 1));
  ret = sum(ret, f(i - 1, x));
  return ret;
}
int main() {
  long long l;
  scanf("%d", &n);
  cin >> l;
  scanf("%d", &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  map<int, int> M2;
  map<int, int> M;
  for (int i = 0; i < n; i++) M2[a[i]]++;
  int j = 0;
  for (map<int, int>::iterator it = M2.begin(); it != M2.end(); it++) {
    b[j] = it->first;
    c[j] = it->second;
    M[it->first] = j;
    j++;
  }
  memset(dp, -1, sizeof dp);
  int ans = 0;
  for (int x = 1; x <= k; x++) {
    long long blocks = l / n;
    if (blocks >= x) {
      int cur = f(j - 1, x);
      int ways = (blocks - x + 1) % 1000000007;
      cur = mul(cur, ways);
      ans = sum(ans, cur);
    }
    if (l % n == 0) {
      continue;
    }
    if (blocks >= x - 1) {
      for (int i = 0; i < l % n; i++) {
        int sn = M[a[i]];
        int cur = f(sn, x - 1);
        ans = sum(ans, cur);
      }
    }
  }
  cout << ans << endl;
}

### Prompt

Please provide a cpp coded solution 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 mul(int a, int b) {
  long long c = a;
  c *= b;
  return c % 1000000007;
}
int sum(int a, int b) {
  a += b;
  return a % 1000000007;
}
int dp[1100000 * 5], a[1100000], b[1100000], c[1100000];
int n, k;
int f(int i, int x) {
  if (i == -1) return x == 0;
  int &ret = dp[i + x * n];
  if (ret != -1) return ret;
  ret = 0;
  if (x) ret = mul(c[i], f(i, x - 1));
  ret = sum(ret, f(i - 1, x));
  return ret;
}
int main() {
  long long l;
  scanf("%d", &n);
  cin >> l;
  scanf("%d", &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  map<int, int> M2;
  map<int, int> M;
  for (int i = 0; i < n; i++) M2[a[i]]++;
  int j = 0;
  for (map<int, int>::iterator it = M2.begin(); it != M2.end(); it++) {
    b[j] = it->first;
    c[j] = it->second;
    M[it->first] = j;
    j++;
  }
  memset(dp, -1, sizeof dp);
  int ans = 0;
  for (int x = 1; x <= k; x++) {
    long long blocks = l / n;
    if (blocks >= x) {
      int cur = f(j - 1, x);
      int ways = (blocks - x + 1) % 1000000007;
      cur = mul(cur, ways);
      ans = sum(ans, cur);
    }
    if (l % n == 0) {
      continue;
    }
    if (blocks >= x - 1) {
      for (int i = 0; i < l % n; i++) {
        int sn = M[a[i]];
        int cur = f(sn, x - 1);
        ans = sum(ans, cur);
      }
    }
  }
  cout << ans << endl;
}

```