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;
template <typename T>
T in() {
  char ch;
  T n = 0;
  bool ng = false;
  while (1) {
    ch = getchar();
    if (ch == '-') {
      ng = true;
      ch = getchar();
      break;
    }
    if (ch >= '0' && ch <= '9') break;
  }
  while (1) {
    n = n * 10 + (ch - '0');
    ch = getchar();
    if (ch < '0' || ch > '9') break;
  }
  return (ng ? -n : n);
}
template <typename T>
inline T Dist(T x1, T y1, T x2, T y2) {
  return sqrt((x1 - x2 * x1 - x2) + (y1 - y2 * y1 - y2));
}
template <typename T>
inline T POW(T B, T P) {
  if (P == 0) return 1;
  if (P & 1)
    return B * POW(B, P - 1);
  else
    return (POW(B, P / 2) * POW(B, P / 2));
}
template <typename T>
inline T ABS(T a) {
  if (a < 0)
    return -a;
  else
    return a;
}
template <typename T>
inline T Gcd(T a, T b) {
  if (a < 0) return Gcd(-a, b);
  if (b < 0) return Gcd(a, -b);
  return (b == 0) ? a : Gcd(b, a % b);
}
template <typename T>
inline T Lcm(T a, T b) {
  if (a < 0) return Lcm(-a, b);
  if (b < 0) return Lcm(a, -b);
  return a * (b / Gcd(a, b));
}
long long Bigmod(long long base, long long power, long long MOD) {
  long long ret = 1;
  while (power) {
    if (power & 1) ret = (ret * base) % MOD;
    base = (base * base) % MOD;
    power >>= 1;
  }
  return ret;
}
long long ModInverse(long long number, long long MOD) {
  return Bigmod(number, MOD - 2, MOD);
}
int Set(int N, int pos) { return N = N | (1 << pos); }
int Reset(int N, int pos) { return N = N & ~(1 << pos); }
bool Chkbit(int N, int pos) { return (bool)(N & (1 << pos)); }
bool isVowel(char ch) {
  ch = toupper(ch);
  if (ch == 'A' || ch == 'U' || ch == 'I' || ch == 'O' || ch == 'E')
    return true;
  return false;
}
bool isConst(char ch) {
  if (isalpha(ch) && !isVowel(ch)) return true;
  return false;
}
int toInt(string s) {
  int sm;
  stringstream ss(s);
  ss >> sm;
  return sm;
}
long long int toLlint(string s) {
  long long int sm;
  stringstream ss(s);
  ss >> sm;
  return sm;
}
long long int a[1000007], b[1000007];
map<long long int, int> mp;
long long int dp[2][1000007], Dp[2][1000007];
int main() {
  long long int n, l, k;
  n = in<long long int>(), l = in<long long int>(), k = in<long long int>();
  long long int tot = l / n;
  long long int baki = l % n;
  for (int i = 1; i <= n; i++) {
    a[i] = in<long long int>();
    b[i] = a[i];
  }
  sort(a + 1, a + n + 1);
  for (int i = n; i >= 1; i--) {
    if (!mp[a[i]]) {
      mp[a[i]] = i;
    }
  }
  dp[0][0] = 1;
  for (int i = 1; i <= n; i++) Dp[0][i] = 1;
  long long int ans = 0;
  for (long long int i = 1; i <= k; i++) {
    int f = i % 2;
    Dp[f][0] = dp[f][0] = 0;
    for (long long int j = 1; j <= n; j++) {
      dp[f][j] = Dp[f ^ 1][mp[a[j]]];
      Dp[f][j] = (Dp[f][j - 1] + dp[f][j]) % 1000000007;
      if (tot >= i) {
        ans = (ans + (dp[f][j] * ((tot - i + 1LL) % 1000000007)) % 1000000007) %
              1000000007;
      }
    }
    for (long long int j = 1; j <= baki; j++) {
      if (tot + 1LL >= i) {
        ans = (ans + Dp[f ^ 1][mp[b[j]]]) % 1000000007;
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}
