#include <bits/stdc++.h>
using namespace std;
long long dp[1000100], sum[1000100];
int n, k;
long long L;
long long getpo(int i, int j) { return ((long long)j - 1) * n + i; }
struct node {
  int x, id;
} no[1000100];
bool cmp(const node& A, const node& B) { return A.x < B.x; }
int a[1000100];
int main() {
  scanf("%d%I64d%d", &n, &L, &k);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    no[i].x = x;
    no[i].id = i;
  }
  sort(no, no + n, cmp);
  for (int i = 0; i < n; i++) {
    int tmp = getpo(i, 1);
    dp[tmp] = 1;
  }
  long long ans = 0;
  long long Left = (L + n - 1) / n;
  for (int j = 2; j <= k; j++) {
    int l = 0, r = 0;
    long long Tmp = 0;
    for (int i = 1; i < n; i++) {
      if (no[i].x != no[i - 1].x) {
        long long num = r - l + 1;
        int now = no[i - 1].id;
        for (int x = l; x <= r; x++) {
          sum[x] = (Tmp + num * dp[getpo(now, j - 1)]) % 1000000007;
        }
        Tmp = (Tmp + num % 1000000007 * dp[getpo(now, j - 1)] % 1000000007) %
              1000000007;
        l = r = i;
      } else
        r = i;
    }
    int num = r - l + 1;
    int now = no[n - 1].id;
    for (int x = l; x <= r; x++)
      sum[x] = (Tmp + num * dp[getpo(now, j - 1)]) % 1000000007;
    for (int i = 0; i < n; i++) {
      int now = no[i].id;
      dp[getpo(now, j)] = sum[i];
    }
  }
  for (int j = 1; j <= k; j++) {
    for (int i = 0; i < n; i++) {
      int now = no[i].id;
      if (getpo(now, j) >= L) continue;
      long long xx = (Left - j + 1);
      if (L % n != 0 && L % n - 1 < now) xx--;
      ans = (ans + xx % 1000000007 * dp[getpo(now, j)]) % 1000000007;
    }
  }
  cout << ans << endl;
  return 0;
}

### 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>
using namespace std;
long long dp[1000100], sum[1000100];
int n, k;
long long L;
long long getpo(int i, int j) { return ((long long)j - 1) * n + i; }
struct node {
  int x, id;
} no[1000100];
bool cmp(const node& A, const node& B) { return A.x < B.x; }
int a[1000100];
int main() {
  scanf("%d%I64d%d", &n, &L, &k);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    no[i].x = x;
    no[i].id = i;
  }
  sort(no, no + n, cmp);
  for (int i = 0; i < n; i++) {
    int tmp = getpo(i, 1);
    dp[tmp] = 1;
  }
  long long ans = 0;
  long long Left = (L + n - 1) / n;
  for (int j = 2; j <= k; j++) {
    int l = 0, r = 0;
    long long Tmp = 0;
    for (int i = 1; i < n; i++) {
      if (no[i].x != no[i - 1].x) {
        long long num = r - l + 1;
        int now = no[i - 1].id;
        for (int x = l; x <= r; x++) {
          sum[x] = (Tmp + num * dp[getpo(now, j - 1)]) % 1000000007;
        }
        Tmp = (Tmp + num % 1000000007 * dp[getpo(now, j - 1)] % 1000000007) %
              1000000007;
        l = r = i;
      } else
        r = i;
    }
    int num = r - l + 1;
    int now = no[n - 1].id;
    for (int x = l; x <= r; x++)
      sum[x] = (Tmp + num * dp[getpo(now, j - 1)]) % 1000000007;
    for (int i = 0; i < n; i++) {
      int now = no[i].id;
      dp[getpo(now, j)] = sum[i];
    }
  }
  for (int j = 1; j <= k; j++) {
    for (int i = 0; i < n; i++) {
      int now = no[i].id;
      if (getpo(now, j) >= L) continue;
      long long xx = (Left - j + 1);
      if (L % n != 0 && L % n - 1 < now) xx--;
      ans = (ans + xx % 1000000007 * dp[getpo(now, j)]) % 1000000007;
    }
  }
  cout << ans << endl;
  return 0;
}

```