#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
const long long mod = 1000000007LL;
vector<long long> dp[maxn], sum[maxn];
int a[maxn], b[maxn], pos[maxn];
long long n, l, k;
int main() {
  scanf("%lld%lld%lld", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  sort(b, b + n);
  for (int i = 0; i < n; i++) {
    pos[i] = upper_bound(b, b + n, b[i]) - b - 1;
    a[i] = upper_bound(b, b + n, a[i]) - b - 1;
  }
  for (int i = 0; i < n; i++) {
    dp[1].push_back(1);
    sum[1].push_back(i + 1);
  }
  for (int i = 2; i <= k && i <= l / n; i++) {
    for (int j = 0; j < n; j++) {
      dp[i].push_back(sum[i - 1][pos[j]]);
      sum[i].push_back(((j ? sum[i][j - 1] : 0) + dp[i][j]) % mod);
    }
  }
  long long ans = 0;
  for (int i = 1; i <= k && i <= l / n; i++) {
    for (int j = 0; j < n; j++) {
      ans += (((l / n - i + 1) % mod) * dp[i][j]) % mod;
      ans %= mod;
    }
  }
  if (l % n) {
    ans += l % n;
    ans %= mod;
    for (int i = 2; i <= k && i <= l / n + 1; i++) {
      for (int j = 0; j < l % n; j++) {
        ans += sum[i - 1][a[j]];
        ans %= mod;
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### 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;
const int maxn = 1e6 + 10;
const long long mod = 1000000007LL;
vector<long long> dp[maxn], sum[maxn];
int a[maxn], b[maxn], pos[maxn];
long long n, l, k;
int main() {
  scanf("%lld%lld%lld", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  sort(b, b + n);
  for (int i = 0; i < n; i++) {
    pos[i] = upper_bound(b, b + n, b[i]) - b - 1;
    a[i] = upper_bound(b, b + n, a[i]) - b - 1;
  }
  for (int i = 0; i < n; i++) {
    dp[1].push_back(1);
    sum[1].push_back(i + 1);
  }
  for (int i = 2; i <= k && i <= l / n; i++) {
    for (int j = 0; j < n; j++) {
      dp[i].push_back(sum[i - 1][pos[j]]);
      sum[i].push_back(((j ? sum[i][j - 1] : 0) + dp[i][j]) % mod);
    }
  }
  long long ans = 0;
  for (int i = 1; i <= k && i <= l / n; i++) {
    for (int j = 0; j < n; j++) {
      ans += (((l / n - i + 1) % mod) * dp[i][j]) % mod;
      ans %= mod;
    }
  }
  if (l % n) {
    ans += l % n;
    ans %= mod;
    for (int i = 2; i <= k && i <= l / n + 1; i++) {
      for (int j = 0; j < l % n; j++) {
        ans += sum[i - 1][a[j]];
        ans %= mod;
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```