#include <bits/stdc++.h>
int dx[4] = {0, -1, 0, 1};
int dy[4] = {1, 0, -1, 0};
using namespace std;
long long n, l, K;
long long a[1234567], ta[1234567], ca[1234567];
long long fuck() {
  sort(ta, ta + n);
  long long ret = 0;
  for (long long i = 0; i < n; i++) {
    long long j = i;
    while (j + 1 < n && ta[j + 1] == ta[i]) j++;
    ta[ret] = ta[i];
    ca[ret] = j - i + 1;
    ret++;
    i = j;
  }
  return ret;
}
vector<long long> dp[1234567];
long long p(long long x, long long m) {
  long long ret = 1;
  x %= 1000000007ll;
  while (m != 0) {
    if (m % 2) ret = ret * x % 1000000007ll;
    m /= 2;
    x = x * x % 1000000007ll;
  }
  return ret;
}
int Main() {
  scanf("%I64d%I64d%I64d", &n, &l, &K);
  for (int i = 0; i < int(n); ++i) scanf("%I64d", &a[i]);
  for (int i = 0; i < int(n); ++i) ta[i] = a[i];
  for (int i = 0; i < int(n + 1); ++i) dp[i].resize(K + 1);
  long long cn = fuck();
  dp[0][0] = 1;
  for (int i = 1; i <= int(cn); ++i)
    for (int j = 0; j < int(K + 1); ++j) {
      if (j == 0) {
        dp[i][j] = 1;
        continue;
      }
      dp[i][j] = 0;
      if (i > 1) dp[i][j] = (dp[i][j] + dp[i - 1][j]) % 1000000007ll;
      dp[i][j] =
          (dp[i][j] + dp[i][j - 1] * ca[i - 1] % 1000000007ll) % 1000000007ll;
    }
  long long ans = 0;
  long long B = l / n;
  for (long long x = 1; x <= K && x <= B; x++) {
    ans = (ans + (B - x + 1) % 1000000007ll * dp[cn][x] % 1000000007ll) %
          1000000007ll;
  }
  long long rn = l % n;
  for (int i = 0; i < int(rn); ++i) {
    long long nn = upper_bound(ta, ta + cn, a[i]) - ta;
    ans = (ans + 1) % 1000000007ll;
    for (long long x = 1; x <= K - 1 && x <= B; x++) {
      ans = (ans + dp[nn][x] % 1000000007ll) % 1000000007ll;
    }
  }
  cout << ans << endl;
  return 0;
}
int main() { return Main(); }

### Prompt

Please create a solution in Cpp 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>
int dx[4] = {0, -1, 0, 1};
int dy[4] = {1, 0, -1, 0};
using namespace std;
long long n, l, K;
long long a[1234567], ta[1234567], ca[1234567];
long long fuck() {
  sort(ta, ta + n);
  long long ret = 0;
  for (long long i = 0; i < n; i++) {
    long long j = i;
    while (j + 1 < n && ta[j + 1] == ta[i]) j++;
    ta[ret] = ta[i];
    ca[ret] = j - i + 1;
    ret++;
    i = j;
  }
  return ret;
}
vector<long long> dp[1234567];
long long p(long long x, long long m) {
  long long ret = 1;
  x %= 1000000007ll;
  while (m != 0) {
    if (m % 2) ret = ret * x % 1000000007ll;
    m /= 2;
    x = x * x % 1000000007ll;
  }
  return ret;
}
int Main() {
  scanf("%I64d%I64d%I64d", &n, &l, &K);
  for (int i = 0; i < int(n); ++i) scanf("%I64d", &a[i]);
  for (int i = 0; i < int(n); ++i) ta[i] = a[i];
  for (int i = 0; i < int(n + 1); ++i) dp[i].resize(K + 1);
  long long cn = fuck();
  dp[0][0] = 1;
  for (int i = 1; i <= int(cn); ++i)
    for (int j = 0; j < int(K + 1); ++j) {
      if (j == 0) {
        dp[i][j] = 1;
        continue;
      }
      dp[i][j] = 0;
      if (i > 1) dp[i][j] = (dp[i][j] + dp[i - 1][j]) % 1000000007ll;
      dp[i][j] =
          (dp[i][j] + dp[i][j - 1] * ca[i - 1] % 1000000007ll) % 1000000007ll;
    }
  long long ans = 0;
  long long B = l / n;
  for (long long x = 1; x <= K && x <= B; x++) {
    ans = (ans + (B - x + 1) % 1000000007ll * dp[cn][x] % 1000000007ll) %
          1000000007ll;
  }
  long long rn = l % n;
  for (int i = 0; i < int(rn); ++i) {
    long long nn = upper_bound(ta, ta + cn, a[i]) - ta;
    ans = (ans + 1) % 1000000007ll;
    for (long long x = 1; x <= K - 1 && x <= B; x++) {
      ans = (ans + dp[nn][x] % 1000000007ll) % 1000000007ll;
    }
  }
  cout << ans << endl;
  return 0;
}
int main() { return Main(); }

```