#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
map<long long, int> m;
long long data[1000000];
int cnt[1000000];
int last[1000000];
int main() {
  int n, k;
  long long l;
  scanf("%d %I64d %d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &data[i]);
    if (m.find(data[i]) != m.end()) {
      m[data[i]]++;
    } else {
      m[data[i]] = 1;
    }
  }
  long long totk = (l - 1) / n + 1;
  if (totk < k) k = totk;
  int s = m.size();
  int i = 0;
  for (auto &x : m) {
    cnt[i] = x.second;
    x.second = i;
    i++;
  }
  int lastlen = l % n;
  if (lastlen == 0) lastlen = n;
  for (i = 0; i < lastlen; i++) {
    last[m[data[i]]]++;
  }
  long long dyn[s][k + 1];
  long long dyn2[s][k + 1];
  for (i = 0; i < s; i++) {
    dyn[i][0] = 1;
    for (int j = 1; j < k + 1; j++) {
      dyn[i][j] = (cnt[i] * dyn[i][j - 1]) % MOD;
      dyn2[i][j] = (last[i] * dyn[i][j - 1]) % MOD;
      if (i > 0) {
        dyn[i][j] = (dyn[i][j] + dyn[i - 1][j]) % MOD;
        dyn2[i][j] = (dyn2[i][j] + dyn2[i - 1][j]) % MOD;
      }
    }
  }
  long long res = 0;
  for (int i = 1; i < k + 1; i++) {
    res = (res + dyn2[s - 1][i]) % MOD;
    res = (res + (((totk - i) % MOD) * dyn[s - 1][i]) % MOD) % MOD;
  }
  printf("%I64d", res);
}

### Prompt

Your challenge is to write 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 = 1000000007;
map<long long, int> m;
long long data[1000000];
int cnt[1000000];
int last[1000000];
int main() {
  int n, k;
  long long l;
  scanf("%d %I64d %d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &data[i]);
    if (m.find(data[i]) != m.end()) {
      m[data[i]]++;
    } else {
      m[data[i]] = 1;
    }
  }
  long long totk = (l - 1) / n + 1;
  if (totk < k) k = totk;
  int s = m.size();
  int i = 0;
  for (auto &x : m) {
    cnt[i] = x.second;
    x.second = i;
    i++;
  }
  int lastlen = l % n;
  if (lastlen == 0) lastlen = n;
  for (i = 0; i < lastlen; i++) {
    last[m[data[i]]]++;
  }
  long long dyn[s][k + 1];
  long long dyn2[s][k + 1];
  for (i = 0; i < s; i++) {
    dyn[i][0] = 1;
    for (int j = 1; j < k + 1; j++) {
      dyn[i][j] = (cnt[i] * dyn[i][j - 1]) % MOD;
      dyn2[i][j] = (last[i] * dyn[i][j - 1]) % MOD;
      if (i > 0) {
        dyn[i][j] = (dyn[i][j] + dyn[i - 1][j]) % MOD;
        dyn2[i][j] = (dyn2[i][j] + dyn2[i - 1][j]) % MOD;
      }
    }
  }
  long long res = 0;
  for (int i = 1; i < k + 1; i++) {
    res = (res + dyn2[s - 1][i]) % MOD;
    res = (res + (((totk - i) % MOD) * dyn[s - 1][i]) % MOD) % MOD;
  }
  printf("%I64d", res);
}

```