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>.
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const int MAX = 1000005;
vector<int> dp[MAX];
int a[MAX];
void add(int &a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
int main() {
  int n, k;
  long long l;
  scanf("%d %I64d %d", &n, &l, &k);
  for (int i = 0; i < n; i++) dp[i].resize(k + 1);
  vector<int> vals;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    vals.push_back(a[i]);
  }
  sort(vals.begin(), vals.end());
  vals.resize(unique(vals.begin(), vals.end()) - vals.begin());
  for (int i = 0; i < n; i++) {
    a[i] = (int)(lower_bound(vals.begin(), vals.end(), a[i]) - vals.begin());
    dp[i][1] = 1;
  }
  vector<int> cnt;
  cnt.resize(vals.size());
  for (int j = 2; j <= k; j++) {
    for (int i = 0; i < (int)cnt.size(); i++) cnt[i] = 0;
    for (int i = 0; i < n; i++) add(cnt[a[i]], dp[i][j - 1]);
    for (int i = 1; i < (int)cnt.size(); i++) add(cnt[i], cnt[i - 1]);
    for (int i = 0; i < n; i++) dp[i][j] = cnt[a[i]];
  }
  int ans = 0;
  for (int i = 0; i < n; i++)
    for (int j = 1; j <= k; j++) {
      long long b = l / n;
      if (i < l % n) b++;
      b = b - j + 1;
      if (b > 0) ans = (ans + (b % MOD) * dp[i][j]) % MOD;
    }
  printf("%d\n", ans);
  return 0;
}
