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;
int n, k;
long long l;
int a[1000100];
int b[1000100];
map<int, int> dp[1000100][2];
map<int, int> dp2[1000100][2];
int sz;
int solveUnbound(int curr, int x, int f) {
  if (x < 0) return 0;
  if (x == 0) return 1;
  if (curr == n) {
    return f;
  }
  if (dp[curr][f].count(x)) {
    return dp[curr][f][x];
  }
  int& ret = dp[curr][f][x];
  ret = 0;
  int pos = lower_bound(a, a + n, a[curr]) - a;
  if (pos < n) {
    ret = (ret % 1000000007 + solveUnbound(pos, x - 1, 1) % 1000000007) %
          1000000007;
  }
  ret = (ret % 1000000007 + solveUnbound(curr + 1, x, f) % 1000000007) %
        1000000007;
  return ret;
}
int solveBounded(int curr, int x, int f) {
  if (x < 0) return 0;
  if (x == 0) return 1;
  if (curr == n) {
    return f;
  }
  if (dp2[curr][f].count(x)) {
    return dp2[curr][f][x];
  }
  int& ret = dp2[curr][f][x];
  ret = 0;
  if (x - 1 > 1) {
    int pos = lower_bound(a, a + n, a[curr]) - a;
    if (pos < n) {
      ret = (ret % 1000000007 + solveBounded(pos, x - 1, 1) % 1000000007) %
            1000000007;
    }
    ret = (ret % 1000000007 + solveBounded(curr + 1, x, f) % 1000000007) %
          1000000007;
  } else if (x - 1 == 1) {
    int pos = lower_bound(b, b + sz, a[curr]) - b;
    if (pos < sz)
      ret = (ret % 1000000007 + (sz - pos) % 1000000007) % 1000000007;
    ret = (ret + 1) % 1000000007;
    ret = (ret % 1000000007 + solveBounded(curr + 1, x, f) % 1000000007) %
          1000000007;
  }
  return ret;
}
inline long long mult(long long a, long long b, long long c) {
  if (a < b) swap(a, b);
  long long r = 0, m = a;
  for (int i = 0; i < 64; ++i) {
    if ((1LL << i) > b) break;
    if (b & 1LL << i) r = (r + m) % c;
    m = (m * 2) % c;
  }
  return r;
}
int main(void) {
  ios ::sync_with_stdio(false);
  cin >> n >> l >> k;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    b[i] = a[i];
  }
  sz = l % n;
  sort(a, a + n);
  sort(b, b + sz);
  k = min((long long)k, l / n + (sz != 0));
  int ans = 0;
  long long ret = solveUnbound(0, k, 0);
  long long tmp = mult(max(0LL, (long long)l / (long long)n - (long long)k + 1),
                       ret, 1000000007);
  ans = (ans % 1000000007 + tmp % 1000000007) % 1000000007;
  if (sz == 0) {
    for (int i = k - 1; i >= 1; --i) {
      ans =
          (ans % 1000000007 + solveUnbound(0, i, 0) % 1000000007) % 1000000007;
    }
  } else if (sz > 0) {
    ans = (ans % 1000000007 + solveBounded(0, k, 0) % 1000000007) % 1000000007;
    for (int i = k - 1; i >= 2; --i) {
      ans =
          (ans % 1000000007 + solveBounded(0, i, 0) % 1000000007) % 1000000007;
    }
    ans = (ans % 1000000007 + sz % 1000000007) % 1000000007;
  }
  cout << ans << "\n";
  return 0;
}
