#include <bits/stdc++.h>
using namespace std;
int n, k;
long long l;
int a[1000005], dp[1000005], c[1000005];
int vis[1000005], tim;
pair<int, int*> b[1000005];
void add(int x, int y) {
  for (; x <= n; x += (x & (-x))) {
    if (vis[x] != tim) vis[x] = tim, c[x] = 0;
    c[x] = (c[x] + y) % 1000000007;
  }
}
int query(int x) {
  int ret = 0;
  for (; x; x -= (x & (-x)))
    if (vis[x] == tim) ret = (ret + c[x]) % 1000000007;
  return ret;
}
int main() {
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    b[i] = make_pair(a[i], a + i);
  }
  sort(b, b + n);
  for (int j = 0, i = 0; i < n; i++) {
    if (i == 0 || b[i].first != b[i - 1].first) j++;
    *b[i].second = j;
  }
  long long tmp = l / n % 1000000007, m = l % n, x = l / n;
  long long ans = l % 1000000007;
  for (int i = 0; i < n; i++) dp[i] = 1;
  for (int i = 1; i < k; i++) {
    tim++;
    for (int j = 0; j < n; j++) add(a[j], dp[j]);
    for (int j = 0; j < n; j++) {
      dp[j] = query(a[j]);
      if (x > i)
        ans = (ans + (tmp - i) % 1000000007 * dp[j] % 1000000007 + 1000000007) %
              1000000007;
      if (tmp >= i && j < m) ans = (ans + dp[j]) % 1000000007;
    }
  }
  cout << ans << endl;
}

### 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;
int n, k;
long long l;
int a[1000005], dp[1000005], c[1000005];
int vis[1000005], tim;
pair<int, int*> b[1000005];
void add(int x, int y) {
  for (; x <= n; x += (x & (-x))) {
    if (vis[x] != tim) vis[x] = tim, c[x] = 0;
    c[x] = (c[x] + y) % 1000000007;
  }
}
int query(int x) {
  int ret = 0;
  for (; x; x -= (x & (-x)))
    if (vis[x] == tim) ret = (ret + c[x]) % 1000000007;
  return ret;
}
int main() {
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    b[i] = make_pair(a[i], a + i);
  }
  sort(b, b + n);
  for (int j = 0, i = 0; i < n; i++) {
    if (i == 0 || b[i].first != b[i - 1].first) j++;
    *b[i].second = j;
  }
  long long tmp = l / n % 1000000007, m = l % n, x = l / n;
  long long ans = l % 1000000007;
  for (int i = 0; i < n; i++) dp[i] = 1;
  for (int i = 1; i < k; i++) {
    tim++;
    for (int j = 0; j < n; j++) add(a[j], dp[j]);
    for (int j = 0; j < n; j++) {
      dp[j] = query(a[j]);
      if (x > i)
        ans = (ans + (tmp - i) % 1000000007 * dp[j] % 1000000007 + 1000000007) %
              1000000007;
      if (tmp >= i && j < m) ans = (ans + dp[j]) % 1000000007;
    }
  }
  cout << ans << endl;
}

```