#include <bits/stdc++.h>
using namespace std;
long long n, m, p, q, t, i, j, k, x, y, z, c, s, d, l, r, l1, l2, lst, a, b, mx,
    cnt, cur, u, v, w;
const long long mod = 1000000007;
vector<pair<long long, long long> > nums;
long long is[1000001];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> l >> k;
  long long dp[k + 1][n];
  for (i = 0; i < k + 1; i++)
    for (j = 0; j < n; j++) dp[i][j] = 0;
  for (i = 0; i < n; i++) {
    cin >> p;
    nums.push_back({p, i});
  }
  long long ans = 0;
  long long segs = l / n;
  for (i = 0; i < l % n; i++) {
    is[i] = 1;
  }
  sort(nums.begin(), nums.end());
  for (j = 0; j < n; j++) {
    dp[1][j] = 1;
    ans = (ans + dp[1][j] * ((segs + (is[nums[j].second])) % mod) % mod) % mod;
  }
  for (i = 2; i <= min(k, segs + (n % l != 0)); i++) {
    long long sm = 0;
    for (j = 0; j < n; j++) {
      sm = (sm + dp[i - 1][j]) % mod;
      dp[i][j] = (dp[i][j] + sm) % mod;
    }
    ans = (ans + dp[i][n - 1] *
                     ((segs - i + 1 + (is[nums[n - 1].second])) % mod) % mod) %
          mod;
    for (j = n - 2; j >= 0; j--) {
      if (nums[j].first == nums[j + 1].first) {
        dp[i][j] = dp[i][j + 1];
      }
      ans = (ans +
             dp[i][j] * ((segs - i + 1 + (is[nums[j].second])) % mod) % mod) %
            mod;
    }
  }
  cout << ans;
}

### 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 n, m, p, q, t, i, j, k, x, y, z, c, s, d, l, r, l1, l2, lst, a, b, mx,
    cnt, cur, u, v, w;
const long long mod = 1000000007;
vector<pair<long long, long long> > nums;
long long is[1000001];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> l >> k;
  long long dp[k + 1][n];
  for (i = 0; i < k + 1; i++)
    for (j = 0; j < n; j++) dp[i][j] = 0;
  for (i = 0; i < n; i++) {
    cin >> p;
    nums.push_back({p, i});
  }
  long long ans = 0;
  long long segs = l / n;
  for (i = 0; i < l % n; i++) {
    is[i] = 1;
  }
  sort(nums.begin(), nums.end());
  for (j = 0; j < n; j++) {
    dp[1][j] = 1;
    ans = (ans + dp[1][j] * ((segs + (is[nums[j].second])) % mod) % mod) % mod;
  }
  for (i = 2; i <= min(k, segs + (n % l != 0)); i++) {
    long long sm = 0;
    for (j = 0; j < n; j++) {
      sm = (sm + dp[i - 1][j]) % mod;
      dp[i][j] = (dp[i][j] + sm) % mod;
    }
    ans = (ans + dp[i][n - 1] *
                     ((segs - i + 1 + (is[nums[n - 1].second])) % mod) % mod) %
          mod;
    for (j = n - 2; j >= 0; j--) {
      if (nums[j].first == nums[j + 1].first) {
        dp[i][j] = dp[i][j + 1];
      }
      ans = (ans +
             dp[i][j] * ((segs - i + 1 + (is[nums[j].second])) % mod) % mod) %
            mod;
    }
  }
  cout << ans;
}

```