#include <bits/stdc++.h>
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int maxn = 1e6 + 5, mod = 1e9 + 7;
int dp[maxn], rec[maxn], num[maxn], tot[maxn], last[maxn], C[maxn], val[maxn],
    remain[maxn], n, k;
long long l;
int main() {
  scanf("%d%I64d%d", &n, &l, &k);
  for (int i = 0; i < n; i++) scanf("%d", val + i + 1);
  int t = 1;
  for (int i = l % n; i; i--) remain[i] = val[i];
  sort(val + 1, val + n + 1);
  for (int i = 1; i <= n;) {
    int res = 0, cmp = val[i];
    while (cmp == val[i]) res++, i++;
    val[t] = cmp;
    num[t++] = res;
  }
  C[0] = 1;
  int ans = 0, _max = min(l / n, 1LL * k);
  for (int j = 1; j <= _max; j++) {
    for (int i = 1; i < t; i++) {
      if (j == 1)
        dp[i] = num[i];
      else
        dp[i] = 1LL * dp[i] * num[i] % mod;
      ans += 1LL * (l / n - j + 1) % mod * dp[i] % mod;
      dp[i] %= mod;
      ans %= mod;
      if (j != k) last[i] = (last[i] + dp[i]) % mod;
      dp[i] += dp[i - 1];
      dp[i] %= mod;
    }
  }
  for (int i = 1; i < t; i++) last[i] = (last[i] + last[i - 1]) % mod;
  for (int i = l % n; i; i--) {
    int k = upper_bound(val + 1, val + t, remain[i]) - val;
    ans = (ans + last[k - 1]) % mod;
  }
  ans = (ans + l % n) % mod;
  printf("%d", ans);
  return 0;
}

### Prompt

Please create a solution in CPP to 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>
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int maxn = 1e6 + 5, mod = 1e9 + 7;
int dp[maxn], rec[maxn], num[maxn], tot[maxn], last[maxn], C[maxn], val[maxn],
    remain[maxn], n, k;
long long l;
int main() {
  scanf("%d%I64d%d", &n, &l, &k);
  for (int i = 0; i < n; i++) scanf("%d", val + i + 1);
  int t = 1;
  for (int i = l % n; i; i--) remain[i] = val[i];
  sort(val + 1, val + n + 1);
  for (int i = 1; i <= n;) {
    int res = 0, cmp = val[i];
    while (cmp == val[i]) res++, i++;
    val[t] = cmp;
    num[t++] = res;
  }
  C[0] = 1;
  int ans = 0, _max = min(l / n, 1LL * k);
  for (int j = 1; j <= _max; j++) {
    for (int i = 1; i < t; i++) {
      if (j == 1)
        dp[i] = num[i];
      else
        dp[i] = 1LL * dp[i] * num[i] % mod;
      ans += 1LL * (l / n - j + 1) % mod * dp[i] % mod;
      dp[i] %= mod;
      ans %= mod;
      if (j != k) last[i] = (last[i] + dp[i]) % mod;
      dp[i] += dp[i - 1];
      dp[i] %= mod;
    }
  }
  for (int i = 1; i < t; i++) last[i] = (last[i] + last[i - 1]) % mod;
  for (int i = l % n; i; i--) {
    int k = upper_bound(val + 1, val + t, remain[i]) - val;
    ans = (ans + last[k - 1]) % mod;
  }
  ans = (ans + l % n) % mod;
  printf("%d", ans);
  return 0;
}

```