#include <bits/stdc++.h>
using namespace std;
long long int dp[2][1000010], s1, s2, ans, l;
long long int g, y;
struct node {
  long long int num, index;
} a[1000010];
long long int cmp(node x, node y) { return x.num < y.num; }
long long int n, k;
int main() {
  scanf("%lld%lld%lld", &n, &l, &k);
  for (long long int i = 0; i < n; i++) {
    scanf("%lld", &a[i].num);
    a[i].index = i;
    dp[0][i] = 1;
  }
  sort(a, a + n, cmp);
  ans = l % 1000000007;
  g = l / n;
  y = l % n;
  k = min(k, (l + n - 1) / n);
  long long int t = 0;
  for (long long int kk = 2; kk <= k; kk++) {
    t = 1 - t;
    dp[t][0] = 0;
    for (long long int ii = 0; ii < n; ii++) {
      if (a[ii].num == a[0].num) {
        dp[t][0] = (dp[t][0] + dp[1 - t][ii]) % 1000000007;
      } else
        break;
    }
    s1 = dp[t][0] % 1000000007;
    s2 = 0;
    if (a[0].index < y) {
      s2 = dp[t][0] % 1000000007;
    }
    for (long long int i = 1; i < n; i++) {
      dp[t][i] = dp[t][i - 1];
      if (a[i].num != a[i - 1].num) {
        for (long long int ii = i; ii < n; ii++) {
          if (a[i].num == a[ii].num) {
            dp[t][i] = (dp[t][i] + dp[1 - t][ii]) % 1000000007;
          } else
            break;
        }
      }
      s1 = (s1 + dp[t][i]) % 1000000007;
      if (a[i].index < y) {
        s2 = (s2 + dp[t][i]) % 1000000007;
      }
    }
    ans += ((g - kk + 1) % 1000000007) * s1 % 1000000007;
    if (ans >= 1000000007) ans -= 1000000007;
    ans += s2;
    if (ans >= 1000000007) ans -= 1000000007;
  }
  ans = (ans + 1000000007) % 1000000007;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Create a solution in Cpp for 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 int dp[2][1000010], s1, s2, ans, l;
long long int g, y;
struct node {
  long long int num, index;
} a[1000010];
long long int cmp(node x, node y) { return x.num < y.num; }
long long int n, k;
int main() {
  scanf("%lld%lld%lld", &n, &l, &k);
  for (long long int i = 0; i < n; i++) {
    scanf("%lld", &a[i].num);
    a[i].index = i;
    dp[0][i] = 1;
  }
  sort(a, a + n, cmp);
  ans = l % 1000000007;
  g = l / n;
  y = l % n;
  k = min(k, (l + n - 1) / n);
  long long int t = 0;
  for (long long int kk = 2; kk <= k; kk++) {
    t = 1 - t;
    dp[t][0] = 0;
    for (long long int ii = 0; ii < n; ii++) {
      if (a[ii].num == a[0].num) {
        dp[t][0] = (dp[t][0] + dp[1 - t][ii]) % 1000000007;
      } else
        break;
    }
    s1 = dp[t][0] % 1000000007;
    s2 = 0;
    if (a[0].index < y) {
      s2 = dp[t][0] % 1000000007;
    }
    for (long long int i = 1; i < n; i++) {
      dp[t][i] = dp[t][i - 1];
      if (a[i].num != a[i - 1].num) {
        for (long long int ii = i; ii < n; ii++) {
          if (a[i].num == a[ii].num) {
            dp[t][i] = (dp[t][i] + dp[1 - t][ii]) % 1000000007;
          } else
            break;
        }
      }
      s1 = (s1 + dp[t][i]) % 1000000007;
      if (a[i].index < y) {
        s2 = (s2 + dp[t][i]) % 1000000007;
      }
    }
    ans += ((g - kk + 1) % 1000000007) * s1 % 1000000007;
    if (ans >= 1000000007) ans -= 1000000007;
    ans += s2;
    if (ans >= 1000000007) ans -= 1000000007;
  }
  ans = (ans + 1000000007) % 1000000007;
  printf("%lld\n", ans);
  return 0;
}

```