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 = 1e9 + 7;
int n, k;
long long l;
vector<int> v;
int get(int x) { return lower_bound(v.begin(), v.end(), x) - v.begin(); }
int main() {
  scanf("%d%lld%d", &n, &l, &k);
  int dp1[n + 1][k + 1], dp2[n + 1][k + 1], sum1[n + 1], sum2[n + 1], a[n + 1];
  memset(dp1, 0, sizeof(dp1));
  memset(dp2, 0, sizeof(dp2));
  memset(sum1, 0, sizeof(sum1));
  memset(sum2, 0, sizeof(sum2));
  memset(a, 0, sizeof(a));
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    v.push_back(a[i]);
  }
  sort(v.begin(), v.end());
  auto it = unique(v.begin(), v.end());
  v.erase(it, v.end());
  int m = v.size();
  for (int i = 0; i < n; i++) {
    dp1[i][k] = 1;
    sum1[get(a[i])] += dp1[i][k];
    sum1[get(a[i])] %= MOD;
  }
  for (int i = 0; i < l % n; i++) {
    dp2[i][k] = 1;
    sum2[get(a[i])] += dp2[i][k];
    sum2[get(a[i])] %= MOD;
  }
  for (int i = m - 1; i >= 0; i--) {
    sum1[i] += sum1[i + 1];
    sum1[i] %= MOD;
    sum2[i] += sum2[i + 1];
    sum2[i] %= MOD;
  }
  for (int j = k - 1; j > 0; j--) {
    for (int i = 0; i < n; i++) {
      dp1[i][j] = sum1[get(a[i])];
      dp2[i][j] = sum2[get(a[i])];
    }
    for (int i = 0; i < m; i++) sum1[i] = sum2[i] = 0;
    for (int i = 0; i < n; i++) {
      sum1[get(a[i])] += dp1[i][j];
      sum1[get(a[i])] %= MOD;
      sum2[get(a[i])] += dp2[i][j];
      sum2[get(a[i])] %= MOD;
    }
    for (int i = m - 1; i >= 0; i--) {
      sum1[i] += sum1[i + 1];
      sum1[i] %= MOD;
      sum2[i] += sum2[i + 1];
      sum2[i] %= MOD;
    }
  }
  int sol = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 1; j <= k; j++) {
      long long v1 = l / n;
      long long v2 = (l + n - 1) / n;
      int cnt = k - j + 1;
      if (cnt <= v1) {
        int res = (v1 - cnt + 1) % MOD;
        res = (res * 1LL * dp1[i][j]) % MOD;
        sol = (sol + res) % MOD;
      }
      if (cnt <= v2) sol = (sol + dp2[i][j]) % MOD;
    }
  }
  printf("%d\n", sol);
  return 0;
}
