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 mult(int x, int y) {
  long long int ans, x1 = (long long int)x, y1 = (long long int)y;
  ans = (x1 * y1) % 1000000007;
  return (int)ans;
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int lcm(int a, int b) { return a * (b / gcd(a, b)); }
long long int pow1(long long int a, long long int b) {
  long long int ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return ans;
}
int mina(int arr[], int n) {
  int x = arr[0], i, pos = 0;
  for (i = 1; i < n; i++) {
    if (arr[i] < x) {
      x = arr[i];
      pos = i;
    }
  }
  return x;
}
int maxa(int arr[], int n) {
  int x = arr[0], i, pos = 0;
  for (i = 1; i < n; i++) {
    if (arr[i] > x) {
      x = arr[i];
      pos = i;
    }
  }
  return x;
}
pair<long long int, int> a[1000005], b[1000005];
vector<vector<long long int> > dp;
map<long long int, int> m1;
int main() {
  long long int n, l, k, i, j, mod, num, k1;
  scanf("%lld", &n);
  scanf("%lld", &l);
  scanf("%lld", &k);
  mod = (l % n);
  dp.resize(k + 2);
  for (i = 0; i < k + 2; i++) dp[i].resize(n + 2);
  num = l / n;
  for (i = 0; i < n; i++) {
    scanf("%lld", &b[i].first);
    m1[b[i].first];
  }
  int cn = 0;
  for (map<long long int, int>::iterator itr = m1.begin(); itr != m1.end();
       itr++) {
    m1[(*itr).first] = cn++;
  }
  for (i = 0; i < n; i++) b[i].second = m1[b[i].first];
  for (i = 0; i < n; i++) dp[1][i] = 1;
  for (i = 2; i < k + 1; i++) {
    long long int cnt[n + 2];
    for (j = 0; j < n; j++) cnt[j] = 0;
    for (j = 0; j < n; j++) {
      cnt[b[j].second] += dp[i - 1][j];
      cnt[b[j].second] %= 1000000007;
    }
    for (j = 1; j < n; j++) {
      cnt[j] += cnt[j - 1];
      cnt[j] %= 1000000007;
    }
    for (j = 0; j < n; j++) dp[i][j] = cnt[b[j].second];
  }
  long long int ans = 0;
  for (i = 0; i < n; i++) {
    for (j = 1; j < k + 1; j++) {
      long long int x1 = num;
      if (i < mod) x1++;
      x1 = (x1 - j + 1) % 1000000007;
      if (x1 > 0) ans += (dp[j][i] * x1) % 1000000007;
      ans %= 1000000007;
    }
  }
  printf("%lld\n", ans);
  return 0;
}
