#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int MAXN = 1e6 + 10;
void add(long long &x, long long y) {
  x += y;
  x %= MOD;
}
pair<int, int> a[MAXN];
int b[MAXN];
long long dp[MAXN];
int main() {
  int n, k;
  long long l;
  while (scanf("%d%I64d%d", &n, &l, &k) != EOF) {
    for (int i = 0; i < n; i++) {
      scanf("%d", &b[i]);
    }
    int m = n * k;
    for (int i = 0; i < m; i++) {
      a[i].first = b[i % n];
      a[i].second = i;
      dp[i] = 0;
    }
    sort(a, a + m);
    long long ans = l % MOD;
    int yu = l % n;
    for (int i = 0; i < m; i++) {
      int id = a[i].second;
      if (id < n) {
        add(dp[id / n], 1);
      } else {
        add(dp[id / n], dp[id / n - 1]);
      }
      if (id < l && id >= n) {
        long long cnt = (l - 1) / n - id / n + 1;
        if (yu && (id % n) > yu - 1) {
          cnt--;
        }
        add(ans, cnt % MOD * dp[id / n - 1] % MOD);
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Develop a solution in CPP 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 MOD = 1e9 + 7;
const int MAXN = 1e6 + 10;
void add(long long &x, long long y) {
  x += y;
  x %= MOD;
}
pair<int, int> a[MAXN];
int b[MAXN];
long long dp[MAXN];
int main() {
  int n, k;
  long long l;
  while (scanf("%d%I64d%d", &n, &l, &k) != EOF) {
    for (int i = 0; i < n; i++) {
      scanf("%d", &b[i]);
    }
    int m = n * k;
    for (int i = 0; i < m; i++) {
      a[i].first = b[i % n];
      a[i].second = i;
      dp[i] = 0;
    }
    sort(a, a + m);
    long long ans = l % MOD;
    int yu = l % n;
    for (int i = 0; i < m; i++) {
      int id = a[i].second;
      if (id < n) {
        add(dp[id / n], 1);
      } else {
        add(dp[id / n], dp[id / n - 1]);
      }
      if (id < l && id >= n) {
        long long cnt = (l - 1) / n - id / n + 1;
        if (yu && (id % n) > yu - 1) {
          cnt--;
        }
        add(ans, cnt % MOD * dp[id / n - 1] % MOD);
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```