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:1677721600")
using namespace std;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double ee = exp(1.0);
const int inf = 0x3f3f3f3f;
const int maxn = 1e6 + 10;
const double pi = acos(-1.0);
const long long iinf = 0x3f3f3f3f3f3f3f3f;
int readT() {
  char c;
  int ret = 0, flg = 0;
  while (c = getchar(), (c < '0' || c > '9') && c != '-')
    ;
  if (c == '-')
    flg = 1;
  else
    ret = c ^ 48;
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c ^ 48);
  return flg ? -ret : ret;
}
long long readTL() {
  char c;
  int flg = 0;
  long long ret = 0;
  while (c = getchar(), (c < '0' || c > '9') && c != '-')
    ;
  if (c == '-')
    flg = 1;
  else
    ret = c ^ 48;
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c ^ 48);
  return flg ? -ret : ret;
}
vector<pair<long long, int> > num;
long long tmp[maxn];
long long arr[maxn];
int bigger[maxn];
long long dp[maxn];
long long sum[maxn];
int main() {
  int n = readT();
  long long l = readTL();
  long long k = readTL();
  for (int i = (0); i <= (n - 1); i++) {
    tmp[i] = readTL();
    arr[i] = tmp[i];
    num.push_back(make_pair(tmp[i], i));
  }
  long long d = l / n;
  long long last = l % n;
  sort(tmp, tmp + last);
  for (int i = (0); i <= (n - 1); i++) {
    int pos = lower_bound(tmp, tmp + last, arr[i]) - tmp;
    bigger[i] = last - pos;
  }
  sort(tmp, tmp + n);
  sort(num.begin(), num.end());
  for (int i = (0); i <= (n - 1); i++) {
    dp[i] = 1;
  }
  sum[0] = dp[num[0].second];
  for (int i = (1); i <= (n - 1); i++) {
    sum[i] = sum[i - 1] + dp[num[i].second];
  }
  for (int i = (1); i <= (k - 1); i++) {
    for (int j = (0); j <= (n - 1); j++) {
      int pos = upper_bound(tmp, tmp + n, arr[j]) - tmp - 1;
      dp[i * n + j] = sum[pos];
    }
    sum[0] = dp[i * n + num[0].second];
    for (int j = (1); j <= (n - 1); j++) {
      sum[j] = (sum[j - 1] + dp[i * n + num[j].second]) % mod;
    }
  }
  k = min(k, d + (last == 0 ? 0 : 1));
  long long ans = l % mod;
  for (int i = 1; i < k; ++i) {
    if (i + 1 <= d) {
      for (int j = 0; j < n; ++j)
        ans = (((d - i) % mod) * ((dp[i * n + j]) % mod) + ans) % mod;
    }
    if (last) {
      if (i <= d) {
        for (int j = 0; j < n; ++j)
          ans = (ans + dp[(i - 1) * n + j] * bigger[j]) % mod;
      }
    }
  }
  cout << ans << endl;
  return 0;
}
