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 int maxn = 1e6 + 10, modd = 1e9 + 7, inf = 0x3f3f3f3f, INF = 0x7fffffff,
          hmod1 = 0x48E2DCE7, hmod2 = 0x60000005;
const int dir[4][2] = {{-1, 0}, {0, -1}, {1, 0}, {0, 1}};
const double eps = 1e-8;
template <class T>
inline void scand(T &x) {
  char c;
  x = 0;
  while ((c = getchar()) < '0')
    ;
  while (c >= '0' && c <= '9') x = x * 10 + (c - 48), c = getchar();
}
inline long long min(long long a, long long b) { return a < b ? a : b; }
inline long long max(long long a, long long b) { return a > b ? a : b; }
inline long long gcd(long long a, long long b) {
  return b == 0 ? a : gcd(b, a % b);
}
inline long long exgcd(long long a, long long b, long long &x, long long &y) {
  long long d;
  (b == 0 ? (x = 1, y = 0, d = a)
          : (d = exgcd(b, a % b, y, x), y -= a / b * x));
  return d;
}
inline long long qpow(long long a, long long n) {
  long long sum = 1;
  while (n) {
    if (n & 1) sum = sum * a % modd;
    a = a * a % modd;
    n >>= 1;
  }
  return sum;
}
inline long long qmul(long long a, long long n) {
  long long sum = 0;
  while (n) {
    if (n & 1) sum = (sum + a) % modd;
    a = (a + a) % modd;
    n >>= 1;
  }
  return sum;
}
inline long long inv(long long a) { return qpow(a, modd - 2); }
inline long long madd(long long a, long long b) {
  return (a % modd + b % modd) % modd;
}
inline long long mmul(long long a, long long b) { return a % modd * b % modd; }
inline void uadd(long long &a, long long b) { a = madd(a, b); }
inline void umul(long long &a, long long b) { a = mmul(a, b); }
long long n, m, l, k, ans, a[maxn], dp[maxn], summ[maxn];
pair<long long, long long> c[maxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n * k; i++) c[i] = make_pair(a[i % n], i);
  sort(c, c + n * k);
  for (int i = 0; i < n * k; i++) {
    if (c[i].second < n)
      dp[c[i].second] = 1;
    else
      dp[c[i].second] = summ[c[i].second / n - 1];
    uadd(summ[c[i].second / n], dp[c[i].second]);
    if (c[i].second < l)
      uadd(ans, mmul(madd((l - c[i].second - 1) / n, 1), dp[c[i].second]));
  }
  cout << ans << endl;
}
