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>
#pragma comment(linker, "/STACK:10240000000,10240000000")
using namespace std;
const int N = 1000010;
const long long mod = 1e9 + 7;
int mp[N] = {0};
int a[N], b[N], aIndex[N], bIndex[N];
long long n, l, k, m;
long long sum[N], now[N], len[N], last[N];
int find(int x) {
  int l = 1, r = m;
  while (r >= l) {
    int mid = (r + l) / 2;
    if (b[mid] == x)
      return mid;
    else if (b[mid] > x)
      r = mid - 1;
    else
      l = mid + 1;
  }
  return -1;
}
int main() {
  cin >> n >> l >> k;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  if (n >= l) {
    cout << l << endl;
    return 0;
  }
  sort(b + 1, b + n + 1);
  m = unique(b + 1, b + n + 1) - (b + 1);
  long long ans = 0;
  memset(len, 0, sizeof(len));
  memset(now, 0, sizeof(now));
  for (int i = 1; i <= n; i++) {
    bIndex[i] = find(b[i]);
    aIndex[i] = find(a[i]);
    mp[aIndex[i]]++;
  }
  for (int j = 1; j <= k; j++) {
    if (j == 1)
      sum[0] = 1;
    else
      sum[0] = 0;
    for (int i = 1; i <= m; i++) {
      sum[i] = (sum[i - 1] + now[i] * mp[bIndex[i]]) % mod;
      now[i] = sum[i];
      len[i] = (len[i] + now[i]) % mod;
    }
    if (j * n > l) {
      int up = l - (j - 1) * n;
      for (int i = 1; i <= up; i++) {
        int in = aIndex[i];
        ans = (ans + len[in]) % mod;
      }
      cout << ans << endl;
      return 0;
    }
    for (int i = 1; i <= n; i++) {
      int in = aIndex[i];
      ans = (ans + last[in]) % mod;
    }
    for (int i = 1; i <= n; i++) {
      last[i] = len[i];
    }
    for (int i = 1; i <= m; i++) {
      ans = (ans + now[i] * mp[bIndex[i]]) % mod;
    }
  }
  l -= n * k;
  long long y = l / n;
  for (int i = 1; i <= m; i++) {
    ans = (ans + y % mod * len[i] % mod * mp[bIndex[i]]) % mod;
  }
  int t = l % n;
  for (int i = 1; i <= t; i++) {
    ans = (ans + len[aIndex[i]]) % mod;
  }
  cout << ans << endl;
  return 0;
}
