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 long long maxn = 1e6 + 5;
const long long mod = 1e9 + 7;
const long long inf = 1e18;
long long n, l, k, a[maxn], b[maxn], ps[maxn];
vector<long long> dp[maxn], vec;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> l >> k;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
    dp[1].push_back(1);
    vec.push_back(a[i]);
  }
  sort(b, b + n);
  sort(vec.begin(), vec.end());
  vec.resize(unique(vec.begin(), vec.end()) - vec.begin());
  long long cur = vec.size() - 1;
  for (long long i = n - 1; i >= 0; i--) {
    while (cur) {
      if (vec[cur - 1] >= b[i])
        cur--;
      else
        break;
    }
    ps[cur]++;
  }
  for (long long i = 1; i < (long long)vec.size(); i++)
    ps[i] = (ps[i] + ps[i - 1]);
  for (long long i = 2; i <= k; i++) {
    for (long long j = 0; j < n; j++) {
      long long x = lower_bound(vec.begin(), vec.end(), a[j]) - vec.begin();
      dp[i].push_back(ps[x]);
    }
    for (long long i = 0; i < (long long)vec.size(); i++) ps[i] = 0;
    for (long long j = 0; j < n; j++) {
      long long x = lower_bound(vec.begin(), vec.end(), a[j]) - vec.begin();
      ps[x] = (ps[x] + dp[i][j]) % mod;
    }
    for (long long i = 1; i < (long long)vec.size(); i++)
      ps[i] = (ps[i] + ps[i - 1]) % mod;
  }
  long long ans = 0, cnt = (l / n);
  for (long long i = 1; i <= k; i++) {
    for (long long j = 0; j < n; j++) {
      if (cnt + 1 < i) continue;
      if ((cnt + 1) > i) {
        ans = (ans + (dp[i][j] * ((cnt - i + 1) % mod) % mod)) % mod;
      }
      if (j >= l % n)
        continue;
      else if (cnt * n != l) {
        ans = (ans + dp[i][j]) % mod;
      }
    }
  }
  cout << ans << endl;
}
